GSEB Class 12 Maths Solutions Chapter 6 Application of Derivatives Exercise 6.3

Question 1. Find the slope of the tangent to the curve \( y = 3x^4 – 4x \) at \( x = 4 \).
Answer: The curve's equation is \( y = 3x^4 - 4x \).
First, we find the derivative of \( y \) with respect to \( x \): \( \frac { dy }{ dx } = 12x^3 – 4 \).
Next, to get the tangent's slope at \( x = 4 \), we substitute \( x = 4 \) into the derivative:
\( \left(\frac { dy }{ dx }\right)_{x=4} = 12 \times 4^3 – 4 \)
\( = 12 \times 64 – 4 \)
\( = 768 – 4 \)
\( = 764 \).
The slope of the tangent at \( x = 4 \) is \( 764 \).
In simple words: To find the slope, first calculate the derivative of the curve's equation. Then, put the given \( x \) value into the derivative to get the exact slope at that specific point.

Exam Tip: Remember that the derivative \( \frac{dy}{dx} \) gives the slope of the tangent line to a curve at any point \( (x, y) \).

Question 2. Find the slope of the tangent to the curve \( y = \frac { x - 1}{ x-2 } \), \( x \neq 2 \) at \( x = 10 \).
Answer: The curve's equation is \( y = \frac{x-1}{x-2} \).
To find the slope, we differentiate \( y \) with respect to \( x \) using the quotient rule \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \):
\( \frac { dy }{ dx } = \frac{1(x-2) - (x-1)1}{(x-2)^2} \)
\( = \frac{x-2 - x+1}{(x-2)^2} \)
\( = \frac{-1}{(x-2)^2} \).
Now, we substitute \( x = 10 \) into the derivative to determine the slope of the tangent:
\( \left(\frac { dy }{ dx }\right)_{x=10} = \frac{-1}{(10-2)^2} \)
\( = \frac{-1}{8^2} \)
\( = \frac{-1}{64} \).
So, the slope of the tangent at \( x = 10 \) is \( \frac{-1}{64} \).
In simple words: First, find the derivative of the function using the division rule. Then, replace \( x \) with \( 10 \) to get the tangent's slope at that spot.

Exam Tip: When using the quotient rule, be careful with signs and make sure to correctly apply the derivative to both the numerator and denominator.

Question 3. Find the slope of the tangent to the curve \( y = x^3 – x + 1 \) at the point whose x-coordinate is 2.
Answer: The equation of the curve is \( y = x^3 – x + 1 \).
To find the slope, we differentiate \( y \) with respect to \( x \):
\( \frac { dy }{ dx } = 3x^2 - 1 \).
Now, we substitute the x-coordinate \( x = 2 \) into the derivative to get the tangent's slope:
\( \left(\frac { dy }{ dx }\right)_{x=2} = 3 \times 2^2 - 1 \)
\( = 3 \times 4 - 1 \)
\( = 12 - 1 \)
\( = 11 \).
The slope of the tangent at \( x = 2 \) is \( 11 \).
In simple words: Take the derivative of the curve's formula. Then, plug in \( x = 2 \) to find the slope of the line touching the curve at that exact point.

Exam Tip: For polynomial functions, differentiating term by term is straightforward. Ensure to apply the power rule correctly for each term.

Question 4. Find the slope of the tangent to the curve \( y = x^3 – 3x + 2 \) at the point whose x-coordinate is 3.
Answer: The curve's equation is \( y = x^3 – 3x + 2 \).
First, we find the derivative of \( y \) with respect to \( x \):
\( \frac { dy }{ dx } = 3x^2 - 3 \).
Now, we substitute the x-coordinate \( x = 3 \) into the derivative to determine the slope of the tangent:
\( \left(\frac { dy }{ dx }\right)_{x=3} = 3 \times 3^2 - 3 \)
\( = 3 \times 9 - 3 \)
\( = 27 - 3 \)
\( = 24 \).
The slope of the tangent at \( x = 3 \) is \( 24 \).
In simple words: Differentiate the curve's equation. Then, put \( x = 3 \) into the derivative to determine the slope of the line that touches the curve at that point.

Exam Tip: Remember that the slope of the tangent is given by the value of the derivative at the specified x-coordinate. Simple substitution is key after differentiation.

Question 5. Find the slope of the normal to the curve \( x = a \cos^3 \theta, y = a \sin^3 \theta \) at \( \theta = \frac { \pi }{ 4 } \).
Answer: The curve's equations are \( x = a \cos^3 \theta \) and \( y = a \sin^3 \theta \).
We differentiate both equations with respect to \( \theta \):
\( \frac { dx }{ d\theta } = a \times 3 \cos^2 \theta (-\sin \theta) = -3a \cos^2 \theta \sin \theta \)
\( \frac { dy }{ d\theta } = a \times 3 \sin^2 \theta (\cos \theta) = 3a \sin^2 \theta \cos \theta \).
Now, we find \( \frac { dy }{ dx } \) using the chain rule:
\( \frac { dy }{ dx } = \frac { dy/d\theta }{ dx/d\theta } = \frac { 3a \sin^2 \theta \cos \theta }{ -3a \cos^2 \theta \sin \theta } \)
\( = - \frac { \sin \theta }{ \cos \theta } \)
\( = - \tan \theta \).
The slope of the tangent at \( \theta = \frac { \pi }{ 4 } \) is:
\( \left(\frac { dy }{ dx }\right)_{\theta = \frac{\pi}{4}} = - \tan \left(\frac { \pi }{ 4 }\right) = -1 \).
The slope of the normal is the negative reciprocal of the tangent's slope:
Slope of normal \( = \frac { -1 }{ \text{slope of tangent} } = \frac { -1 }{ -1 } = 1 \).
In simple words: First, find the derivatives of \( x \) and \( y \) with respect to \( \theta \). Then use these to get \( \frac{dy}{dx} \). Plug in \( \theta = \frac{\pi}{4} \) to get the tangent's slope. Finally, flip the tangent's slope and change its sign to get the normal's slope.

Exam Tip: When dealing with parametric equations, remember to use the chain rule to find \( \frac{dy}{dx} \) as \( \frac{dy/d\theta}{dx/d\theta} \). The slope of the normal is always \( -1 \) divided by the slope of the tangent.

Question 6. Find the slope of the normal to the curve \( x = 1 - a \sin \theta, y = b \cos^2 \theta \) at \( \theta = \frac { \pi }{ 2 } \).
Answer: The equation of the curve is \( x = 1 - a \sin \theta \) and \( y = b \cos^2 \theta \).
First, we differentiate both equations with respect to \( \theta \):
\( \frac { dx }{ d\theta } = -a \cos \theta \)
\( \frac { dy }{ d\theta } = b \times 2 \cos \theta (-\sin \theta) = -2b \sin \theta \cos \theta \).
Next, we find \( \frac { dy }{ dx } \) using the chain rule:
\( \frac { dy }{ dx } = \frac { dy/d\theta }{ dx/d\theta } = \frac { -2b \sin \theta \cos \theta }{ -a \cos \theta } = \frac { 2b }{ a } \sin \theta \).
Now, we evaluate the slope of the tangent at \( \theta = \frac { \pi }{ 2 } \):
\( \left(\frac { dy }{ dx }\right)_{\theta = \frac{\pi}{2}} = \frac { 2b }{ a } \sin \left(\frac { \pi }{ 2 }\right) = \frac { 2b }{ a } \times 1 = \frac { 2b }{ a } \).
The slope of the normal is the negative reciprocal of the tangent's slope:
Slope of normal \( = \frac { -1 }{ \frac{2b}{a} } = \frac { -a }{ 2b } \).
In simple words: Calculate the derivatives of \( x \) and \( y \) by \( \theta \). Use these to find \( \frac{dy}{dx} \). Then, put \( \theta = \frac{\pi}{2} \) into \( \frac{dy}{dx} \) for the tangent's slope. Finally, flip that slope and change its sign for the normal's slope.

Exam Tip: Be cautious with trigonometric derivatives and their values at common angles like \( \frac{\pi}{2} \). Also, recall that \( \sin(\frac{\pi}{2}) = 1 \) and \( \cos(\frac{\pi}{2}) = 0 \).

Question 7. Find points at which the tangent to the curve \( y = x^3 – 3x^2 – 9x + 7 \) is parallel to the x-axis.
Answer: The equation of the curve is \( y = x^3 – 3x^2 – 9x + 7 \).
First, we find the derivative of \( y \) with respect to \( x \):
\( \frac { dy }{ dx } = 3x^2 - 6x – 9 \).
We can factor this expression: \( \frac { dy }{ dx } = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) \).
If the tangent is parallel to the x-axis, its slope must be \( 0 \). So, we set the derivative to \( 0 \):
\( 3(x - 3)(x + 1) = 0 \).
This means either \( x - 3 = 0 \) or \( x + 1 = 0 \), giving \( x = 3 \) or \( x = -1 \).
Now we find the corresponding y-coordinates for these x-values:
When \( x = -1 \):
\( y = (-1)^3 – 3(-1)^2 – 9(-1) + 7 \)
\( = -1 - 3(1) + 9 + 7 \)
\( = -1 - 3 + 9 + 7 = 12 \).
So, one point is \( (-1, 12) \).
When \( x = 3 \):
\( y = (3)^3 – 3(3)^2 – 9(3) + 7 \)
\( = 27 - 3(9) - 27 + 7 \)
\( = 27 - 27 - 27 + 7 = -20 \).
So, the other point is \( (3, -20) \).
Therefore, the tangents to the curve are parallel to the x-axis at the points \( (-1, 12) \) and \( (3, -20) \).
In simple words: Find the derivative of the curve. Set it to zero because a tangent parallel to the x-axis has a slope of zero. Solve for \( x \), then use those \( x \) values to find the matching \( y \) values from the original curve equation.

Exam Tip: Tangents parallel to the x-axis have a slope of zero, meaning \( \frac{dy}{dx} = 0 \). Always remember to find both x and y coordinates of the points.

Question 8. Find a point on the curve \( y = (x - 2)^2 \) at which the tangent is parallel to the chord joining the points \( (2, 0) \) and \( (4, 4) \).
Answer: The equation of the curve is \( y = (x - 2)^2 \).
First, we differentiate \( y \) with respect to \( x \) to find the slope of the tangent:
\( \frac { dy }{ dx } = 2(x - 2) \).
Next, we determine the slope of the chord connecting points \( A(2, 0) \) and \( B(4, 4) \).
Slope of chord AB \( = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 0}{4 - 2} = \frac{4}{2} = 2 \).
Since the tangent is parallel to the chord, their slopes must be equal.
So, we set the slope of the tangent equal to the slope of the chord:
\( 2(x - 2) = 2 \)
\( x - 2 = 1 \)
\( x = 3 \).
Now, we find the corresponding y-coordinate on the curve for \( x = 3 \):
\( y = (3 - 2)^2 = 1^2 = 1 \).
Therefore, the point on the curve where the tangent is parallel to the chord AB is \( (3, 1) \).
In simple words: Find the slope of the tangent by taking the derivative. Find the slope of the line connecting the two given points. Set these two slopes equal to each other to find the \( x \)-coordinate. Finally, plug that \( x \)-coordinate back into the original curve's equation to find the \( y \)-coordinate of the point.

Exam Tip: Remember that parallel lines have equal slopes. The slope of a chord is found using the two given points, and the slope of the tangent is found using the derivative.

Question 9. Find the point on the curve \( y = x^3 – 11x + 5 \) at which the tangent is \( y = x - 11 \).
Answer: The equation of the curve is \( y = x^3 – 11x + 5 \).
First, we find the derivative of \( y \) with respect to \( x \) to get the slope of the tangent:
\( \frac { dy }{ dx } = 3x^2 - 11 \). (Equation 1)
The given tangent line is \( y = x - 11 \). The slope of this line is the coefficient of \( x \), which is \( 1 \). (Equation 2)
For the tangent to be the given line, their slopes must be equal.
Equating (1) and (2):
\( 3x^2 - 11 = 1 \)
\( 3x^2 = 12 \)
\( x^2 = 4 \)
\( x = \pm 2 \).
Now we find the corresponding y-coordinates on the curve for these x-values:
When \( x = 2 \):
\( y = (2)^3 – 11(2) + 5 = 8 - 22 + 5 = -9 \).
This gives the point \( (2, -9) \).
When \( x = -2 \):
\( y = (-2)^3 – 11(-2) + 5 = -8 + 22 + 5 = 19 \).
This gives the point \( (-2, 19) \).
Now, we check which of these points lies on the given tangent line \( y = x - 11 \):
For point \( (2, -9) \): Substitute \( x = 2 \) into \( y = x - 11 \): \( y = 2 - 11 = -9 \). This matches, so \( (2, -9) \) is the point.
For point \( (-2, 19) \): Substitute \( x = -2 \) into \( y = x - 11 \): \( y = -2 - 11 = -13 \). This does not match \( 19 \), so \( (-2, 19) \) is not on the given tangent line.
Therefore, the point on the curve where the tangent is \( y = x - 11 \) is \( (2, -9) \).
In simple words: First, find the derivative of the curve's equation to get the slope of any tangent. Then, find the slope of the given tangent line. Set these two slopes equal and solve for \( x \). Use these \( x \) values to find the \( y \) values on the original curve. Finally, check which of these points also lies on the given tangent line.

Exam Tip: After finding potential points, always verify that the point lies on *both* the original curve and the tangent line equation. This confirms it is the correct point of tangency.

Question 10. Find the equations of all lines having slope \( -1 \) that are tangents to the curve \( y = \frac { 1 }{ x-1 } \), \( x \neq 1 \).
Answer: The equation of the curve is \( y = \frac { 1 }{ x-1 } \).
First, we find the derivative of \( y \) with respect to \( x \) to get the slope of the tangent:
\( \frac { dy }{ dx } = -1(x-1)^{-2} = \frac{-1}{(x-1)^2} \).
We are given that the slope of the tangent is \( -1 \). So, we set the derivative equal to \( -1 \):
\( \frac{-1}{(x-1)^2} = -1 \)
\( (x-1)^2 = 1 \).
Taking the square root of both sides, \( x-1 = \pm 1 \).
This gives two possible values for \( x \):
\( x - 1 = 1 \implies x = 2 \)
\( x - 1 = -1 \implies x = 0 \).
Now, we find the corresponding y-coordinates on the curve for these x-values:
When \( x = 2 \): \( y = \frac { 1 }{ 2-1 } = 1 \). So, point \( (2, 1) \).
When \( x = 0 \): \( y = \frac { 1 }{ 0-1 } = -1 \). So, point \( (0, -1) \).
Now, we find the equation of the tangent line for each point using \( y - y_1 = m(x - x_1) \) with \( m = -1 \):
For point \( (2, 1) \):
\( y - 1 = -1(x - 2) \)
\( y - 1 = -x + 2 \)
\( x + y - 3 = 0 \).
For point \( (0, -1) \):
\( y - (-1) = -1(x - 0) \)
\( y + 1 = -x \)
\( x + y + 1 = 0 \).
Therefore, the required tangent lines are \( x + y - 3 = 0 \) and \( x + y + 1 = 0 \).
In simple words: Get the curve's derivative to find the tangent's slope. Set this slope to \( -1 \) and solve for \( x \). Use these \( x \) values to find the \( y \) values on the curve. Then, for each point, use the point-slope form with slope \( -1 \) to write the equation of the tangent line.

Exam Tip: Remember that a quadratic equation like \( (x-1)^2 = 1 \) will generally yield two solutions for \( x \), leading to two distinct points of tangency and thus two tangent line equations.

Question 11. Find the equations of all lines having slope \( 2 \) and that are tangents to the curve \( y = \frac { 1 }{ x-3 } \).
Answer: The equation of the curve is \( y = \frac { 1 }{ x-3 } \).
First, we find the derivative of \( y \) with respect to \( x \) to get the slope of the tangent:
\( \frac { dy }{ dx } = -1(x-3)^{-2} = \frac{-1}{(x-3)^2} \).
We are given that the slope of the tangent is \( 2 \). So, we set the derivative equal to \( 2 \):
\( \frac{-1}{(x-3)^2} = 2 \)
\( -1 = 2(x-3)^2 \)
\( (x-3)^2 = -\frac{1}{2} \).
Since the square of any real number cannot be negative, there is no real value of \( x \) that satisfies this equation. Therefore, there are no points on the curve where the tangent has a slope of \( 2 \).
This means there are no such tangent lines to the given curve with a slope of \( 2 \).
In simple words: Calculate the derivative of the curve. Set this derivative equal to the given slope, \( 2 \). Try to solve for \( x \). If the equation leads to a square equaling a negative number, it means no real solution for \( x \) exists, so there are no such tangent lines.

Exam Tip: Be aware that not every given slope will result in a real tangent line. If solving for \( x \) leads to non-real solutions (like taking the square root of a negative number), then no such tangent exists.

Question 12. Find the equations of all lines having slope \( 0 \) and that are tangents to the curve \( y = \frac{1}{x^{2}-2x+3} \).
Answer: The equation of the curve is \( y = \frac{1}{x^2-2x+3} \).
Let \( (x_1, y_1) \) be the point of tangency.
First, we find the derivative of \( y \) with respect to \( x \) using the chain rule (or quotient rule):
\( \frac { dy }{ dx } = -1(x^2-2x+3)^{-2}(2x-2) \)
\( = - \frac{2x-2}{(x^2-2x+3)^2} \)
\( = - \frac{2(x-1)}{(x^2-2x+3)^2} \).
We are given that the slope of the tangent is \( 0 \). So, we set the derivative equal to \( 0 \):
\( - \frac{2(x_1-1)}{(x_1^2-2x_1+3)^2} = 0 \).
For this fraction to be zero, the numerator must be zero (and the denominator non-zero).
\( 2(x_1-1) = 0 \)
\( x_1 - 1 = 0 \)
\( x_1 = 1 \).
Now, we find the corresponding y-coordinate on the curve for \( x_1 = 1 \):
\( y_1 = \frac{1}{(1)^2 - 2(1) + 3} = \frac{1}{1 - 2 + 3} = \frac{1}{2} \).
So, the point of tangency is \( \left(1, \frac{1}{2}\right) \).
Now, we find the equation of the tangent line using \( y - y_1 = m(x - x_1) \) with \( m = 0 \):
\( y - \frac{1}{2} = 0(x - 1) \)
\( y - \frac{1}{2} = 0 \)
\( y = \frac{1}{2} \).
Therefore, the equation of the tangent line is \( y = \frac{1}{2} \).
In simple words: Take the derivative of the curve's formula. Set it equal to \( 0 \) to find the \( x \) value where the tangent is flat. Use this \( x \) value in the original curve's formula to find the \( y \) value. Then, write the equation of the line using this point and a slope of \( 0 \).

Exam Tip: A slope of zero indicates a horizontal tangent line. The equation of a horizontal line is always of the form \( y = \text{constant} \).

Question 13. Find the points on the curve \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \) at which the tangents are (i) parallel to x-axis, (ii) parallel to y-axis.
Answer: The equation of the curve is \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \).
First, we differentiate both sides with respect to \( x \):
\( \frac{2x}{9} + \frac{2y}{16} \frac{dy}{dx} = 0 \).
To solve for \( \frac{dy}{dx} \):
\( \frac{2y}{16} \frac{dy}{dx} = -\frac{2x}{9} \)
\( \frac{y}{8} \frac{dy}{dx} = -\frac{2x}{9} \)
\( \frac{dy}{dx} = -\frac{2x}{9} \times \frac{8}{y} \)
\( \frac{dy}{dx} = -\frac{16x}{9y} \).

(i) If the tangent is parallel to the x-axis, its slope \( \frac{dy}{dx} \) must be \( 0 \).
\( -\frac{16x}{9y} = 0 \).
This implies that the numerator \( -16x \) must be \( 0 \), so \( x = 0 \). (Note: \( y \neq 0 \) for the derivative to be defined).
Substitute \( x = 0 \) into the original curve equation:
\( \frac{0^2}{9} + \frac{y^2}{16} = 1 \)
\( \frac{y^2}{16} = 1 \)
\( y^2 = 16 \implies y = \pm 4 \).
Thus, the points where the tangents are parallel to the x-axis are \( (0, 4) \) and \( (0, -4) \).

(ii) If the tangent is parallel to the y-axis, its slope \( \frac{dy}{dx} \) must be undefined. This occurs when the denominator of \( \frac{dy}{dx} \) is \( 0 \).
\( 9y = 0 \implies y = 0 \).
Substitute \( y = 0 \) into the original curve equation:
\( \frac{x^2}{9} + \frac{0^2}{16} = 1 \)
\( \frac{x^2}{9} = 1 \)
\( x^2 = 9 \implies x = \pm 3 \).
Thus, the points where the tangents are parallel to the y-axis are \( (3, 0) \) and \( (-3, 0) \).
In simple words: First, find the derivative of the curve's equation. For tangents parallel to the x-axis, set the derivative to zero and solve for \( x \), then find the corresponding \( y \) values. For tangents parallel to the y-axis, set the derivative's denominator to zero and solve for \( y \), then find the corresponding \( x \) values.

Exam Tip: Remember that a tangent parallel to the x-axis has a slope of 0, while a tangent parallel to the y-axis has an undefined slope (the denominator of \( \frac{dy}{dx} \) is 0).

Question 14. Find the equations of the tangent and normal to the given curves at the indicated points :
(i) \( y = x^4 – 6x^3 + 13x^2 – 10x + 5 \) at \( (0, 5) \).
(ii) \( y = x^4 – 6x^3 + 13x^2 – 10x + 5 \) at \( (1, 3) \).
(iii) \( y = x^3 \) at \( (1,1) \).
(iv) \( y = x^2 \) at \( (0, 0) \)
(v) \( x = \cos t, y = \sin t \) at \( t = \frac { \pi }{ 4 } \).
Answer:
(i) We have the curve \( y = x^4 – 6x^3 + 13x^2 – 10x + 5 \). The point is \( (0, 5) \).
First, find the derivative:
\( \frac { dy }{ dx } = 4x^3 – 18x^2 + 26x – 10 \).
Now, evaluate the slope of the tangent at \( x = 0 \):
\( \left(\frac { dy }{ dx }\right)_{x=0} = 4(0)^3 – 18(0)^2 + 26(0) – 10 = -10 \).
The slope of the tangent at \( (0, 5) \) is \( -10 \).
Equation of the tangent at \( (0, 5) \) using \( y - y_1 = m(x - x_1) \):
\( y - 5 = -10(x - 0) \)
\( y - 5 = -10x \)
\( 10x + y - 5 = 0 \).
The slope of the normal is \( - \frac{1}{\text{slope of tangent}} = - \frac{1}{-10} = \frac{1}{10} \).
Equation of the normal at \( (0, 5) \) using \( (x - x_1) + m_t(y - y_1) = 0 \):
\( (x - 0) + (-10)(y - 5) = 0 \)
\( x - 10y + 50 = 0 \).
In simple words: For part (i), differentiate the curve to find the tangent's slope at \( x=0 \). Use this slope and the point \( (0,5) \) to write the tangent's equation. Then, use the negative reciprocal of that slope to find the normal's equation.

(ii) We have the curve \( y = x^4 – 6x^3 + 13x^2 – 10x + 5 \). The point is \( (1, 3) \).
The derivative is \( \frac { dy }{ dx } = 4x^3 – 18x^2 + 26x – 10 \).
Now, evaluate the slope of the tangent at \( x = 1 \):
\( \left(\frac { dy }{ dx }\right)_{x=1} = 4(1)^3 – 18(1)^2 + 26(1) – 10 = 4 - 18 + 26 - 10 = 2 \).
The slope of the tangent at \( (1, 3) \) is \( 2 \).
Equation of the tangent at \( (1, 3) \):
\( y - 3 = 2(x - 1) \)
\( y - 3 = 2x - 2 \)
\( y = 2x + 1 \).
The slope of the normal is \( - \frac{1}{\text{slope of tangent}} = - \frac{1}{2} \).
Equation of the normal at \( (1, 3) \) using \( (x - x_1) + m_t(y - y_1) = 0 \):
\( (x - 1) + 2(y - 3) = 0 \)
\( x - 1 + 2y - 6 = 0 \)
\( x + 2y - 7 = 0 \).
In simple words: For part (ii), differentiate the curve to find the tangent's slope at \( x=1 \). Use this slope and the point \( (1,3) \) to write the tangent's equation. Then, use the negative reciprocal of that slope to find the normal's equation.

(iii) We have the curve \( y = x^3 \). The point is \( (1, 1) \).
First, find the derivative:
\( \frac { dy }{ dx } = 3x^2 \).
Now, evaluate the slope of the tangent at \( x = 1 \):
\( \left(\frac { dy }{ dx }\right)_{x=1} = 3(1)^2 = 3 \).
The slope of the tangent at \( (1, 1) \) is \( 3 \).
Equation of the tangent at \( (1, 1) \):
\( y - 1 = 3(x - 1) \)
\( y - 1 = 3x - 3 \)
\( y = 3x - 2 \).
The slope of the normal is \( - \frac{1}{\text{slope of tangent}} = - \frac{1}{3} \).
Equation of the normal at \( (1, 1) \) using \( (x - x_1) + m_t(y - y_1) = 0 \):
\( (x - 1) + 3(y - 1) = 0 \)
\( x - 1 + 3y - 3 = 0 \)
\( x + 3y - 4 = 0 \).
In simple words: For part (iii), differentiate the curve to find the tangent's slope at \( x=1 \). Use this slope and the point \( (1,1) \) to write the tangent's equation. Then, use the negative reciprocal of that slope to find the normal's equation.

(iv) We have the curve \( y = x^2 \). The point is \( (0, 0) \).
First, find the derivative:
\( \frac { dy }{ dx } = 2x \).
Now, evaluate the slope of the tangent at \( x = 0 \):
\( \left(\frac { dy }{ dx }\right)_{x=0} = 2(0) = 0 \).
The slope of the tangent at \( (0, 0) \) is \( 0 \).
Equation of the tangent at \( (0, 0) \):
\( y - 0 = 0(x - 0) \)
\( y = 0 \).
The slope of the normal is \( - \frac{1}{\text{slope of tangent}} = - \frac{1}{0} \), which is undefined. This means the normal is a vertical line.
Equation of the normal at \( (0, 0) \) using \( (x - x_1) + m_t(y - y_1) = 0 \):
\( (x - 0) + 0(y - 0) = 0 \)
\( x = 0 \).
In simple words: For part (iv), differentiate the curve to find the tangent's slope at \( x=0 \). Use this slope and the point \( (0,0) \) to write the tangent's equation. Since the tangent's slope is zero, the normal will be a vertical line, and its equation will be \( x = 0 \).

(v) We have the parametric equations \( x = \cos t \) and \( y = \sin t \). The point is at \( t = \frac { \pi }{ 4 } \).
First, find the coordinates of the point at \( t = \frac { \pi }{ 4 } \):
\( x = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
\( y = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \).
So, the point is \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \).
Next, differentiate \( x \) and \( y \) with respect to \( t \):
\( \frac { dx }{ dt } = -\sin t \)
\( \frac { dy }{ dt } = \cos t \).
Now, find \( \frac { dy }{ dx } = \frac { dy/dt }{ dx/dt } = \frac { \cos t }{ -\sin t } = -\cot t \).
Evaluate the slope of the tangent at \( t = \frac { \pi }{ 4 } \):
\( \left(\frac { dy }{ dx }\right)_{t=\frac{\pi}{4}} = -\cot\left(\frac{\pi}{4}\right) = -1 \).
The slope of the tangent is \( -1 \).
Equation of the tangent at \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \):
\( y - \frac{1}{\sqrt{2}} = -1\left(x - \frac{1}{\sqrt{2}}\right) \)
\( y - \frac{1}{\sqrt{2}} = -x + \frac{1}{\sqrt{2}} \)
\( x + y - \frac{2}{\sqrt{2}} = 0 \)
\( x + y - \sqrt{2} = 0 \).
The slope of the normal is \( - \frac{1}{\text{slope of tangent}} = - \frac{1}{-1} = 1 \).
Equation of the normal at \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \):
\( y - \frac{1}{\sqrt{2}} = 1\left(x - \frac{1}{\sqrt{2}}\right) \)
\( y = x \).
In simple words: For part (v), first find the \( x \) and \( y \) coordinates using \( t = \frac{\pi}{4} \). Then, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), and use them to calculate \( \frac{dy}{dx} \). Plug in \( t = \frac{\pi}{4} \) to get the tangent's slope. Write the tangent's equation using this slope and the point. Finally, calculate the negative reciprocal of the tangent's slope for the normal's equation.

Exam Tip: For parametric equations, convert the parameter to x and y coordinates first if needed, then use \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Always remember to simplify expressions involving square roots.

Question 15. Find the equation of the tangent line to the curve \( y = x^2 – 2x + 7 \) which is
(a) parallel to the line \( 2x – y + 9 = 0 \).
(b) perpendicular to the line \( 5y – 15x = 13 \).
Answer: The equation of the curve is \( y = x^2 – 2x + 7 \).
First, we find the derivative of \( y \) with respect to \( x \) to get the slope of the tangent:
\( \frac { dy }{ dx } = 2x - 2 = 2(x - 1) \).

(a) The tangent is parallel to the line \( 2x – y + 9 = 0 \).
We rearrange the given line equation to find its slope: \( y = 2x + 9 \).
The slope of this line is \( 2 \).
Since the tangent is parallel to this line, the slope of the tangent must also be \( 2 \).
Set \( \frac { dy }{ dx } = 2 \):
\( 2(x - 1) = 2 \)
\( x - 1 = 1 \)
\( x = 2 \).
Now, find the corresponding y-coordinate on the curve for \( x = 2 \):
\( y = (2)^2 – 2(2) + 7 = 4 - 4 + 7 = 7 \).
So, the point of tangency is \( (2, 7) \).
Equation of the tangent line at \( (2, 7) \) with slope \( 2 \):
\( y - 7 = 2(x - 2) \)
\( y - 7 = 2x - 4 \)
\( 2x - y + 3 = 0 \).
In simple words: For part (a), find the derivative of the curve. Find the slope of the given parallel line. Set the derivative equal to this slope to find \( x \). Use this \( x \) to find \( y \) on the curve. Then, write the tangent's equation using the point and slope.

(b) The tangent is perpendicular to the line \( 5y – 15x = 13 \).
We rearrange the given line equation to find its slope: \( 5y = 15x + 13 \implies y = 3x + \frac{13}{5} \).
The slope of this line is \( 3 \).
Since the tangent is perpendicular to this line, the slope of the tangent must be the negative reciprocal of \( 3 \), which is \( -\frac{1}{3} \).
Set \( \frac { dy }{ dx } = -\frac{1}{3} \):
\( 2(x - 1) = -\frac{1}{3} \)
\( x - 1 = -\frac{1}{6} \)
\( x = 1 - \frac{1}{6} = \frac{6-1}{6} = \frac{5}{6} \).
Now, find the corresponding y-coordinate on the curve for \( x = \frac{5}{6} \):
\( y = \left(\frac{5}{6}\right)^2 - 2\left(\frac{5}{6}\right) + 7 \)
\( y = \frac{25}{36} - \frac{10}{6} + 7 \)
\( y = \frac{25}{36} - \frac{60}{36} + \frac{252}{36} \)
\( y = \frac{25 - 60 + 252}{36} = \frac{217}{36} \).
So, the point of tangency is \( \left(\frac{5}{6}, \frac{217}{36}\right) \).
Equation of the tangent line at \( \left(\frac{5}{6}, \frac{217}{36}\right) \) with slope \( -\frac{1}{3} \):
\( y - \frac{217}{36} = -\frac{1}{3}\left(x - \frac{5}{6}\right) \)
Multiply by 36 to clear denominators:
\( 36\left(y - \frac{217}{36}\right) = 36\left(-\frac{1}{3}\right)\left(x - \frac{5}{6}\right) \)
\( 36y - 217 = -12\left(x - \frac{5}{6}\right) \)
\( 36y - 217 = -12x + 10 \)
\( 12x + 36y - 227 = 0 \).
In simple words: For part (b), find the derivative of the curve. Find the slope of the given perpendicular line, then take its negative reciprocal to get the tangent's slope. Set the derivative equal to this reciprocal slope to find \( x \). Use this \( x \) to find \( y \) on the curve. Then, write the tangent's equation using the point and the tangent's slope.

Exam Tip: Recall that parallel lines have equal slopes, and perpendicular lines have slopes that are negative reciprocals of each other (i.e., \( m_1 m_2 = -1 \)). Always simplify your final equation to the standard form \( Ax + By + C = 0 \).

Question 16. Show that the tangents to the curve \( y = 7x^3 + 11 \) at the points, where \( x = 2 \) and \( x = -2 \), are parallel.
Answer: The equation of the curve is \( y = 7x^3 + 11 \).
First, we find the derivative of \( y \) with respect to \( x \) to get the general slope of the tangent:
\( \frac { dy }{ dx } = 7 \times 3x^2 = 21x^2 \).
Now, we evaluate the slope of the tangent at the given x-coordinates:
At \( x = 2 \):
Slope \( m_1 = 21(2)^2 = 21 \times 4 = 84 \).
At \( x = -2 \):
Slope \( m_2 = 21(-2)^2 = 21 \times 4 = 84 \).
Since \( m_1 = m_2 = 84 \), the slopes of the tangents at \( x = 2 \) and \( x = -2 \) are equal. Therefore, the tangents to the given curve at these points are parallel.
In simple words: Differentiate the curve's equation to find the formula for the tangent's slope. Calculate this slope for \( x=2 \) and for \( x=-2 \). If both slopes are the same, the tangents are parallel.

Exam Tip: To show that two lines are parallel, you must demonstrate that their slopes are identical. The derivative provides the instantaneous slope at any point on the curve.

Question 17. Find the points on the curve \( y = x^3 \) at which the slope of the tangent is equal to the y-coordinate of the point.
Answer: The equation of the curve is \( y = x^3 \).
Let \( (x_1, y_1) \) be a point on the curve. So, \( y_1 = x_1^3 \). (Equation 1)
First, we find the derivative of \( y \) with respect to \( x \) to get the slope of the tangent:
\( \frac { dy }{ dx } = 3x^2 \).
The slope of the tangent at the point \( (x_1, y_1) \) is \( 3x_1^2 \).
According to the problem statement, the slope of the tangent is equal to the y-coordinate of the point. So:
\( 3x_1^2 = y_1 \). (Equation 2)
Now, we substitute Equation 1 into Equation 2:
\( 3x_1^2 = x_1^3 \).
Rearrange the equation to solve for \( x_1 \):
\( x_1^3 - 3x_1^2 = 0 \)
\( x_1^2(x_1 - 3) = 0 \).
This equation gives two possible values for \( x_1 \):
\( x_1^2 = 0 \implies x_1 = 0 \)
\( x_1 - 3 = 0 \implies x_1 = 3 \).
Now, we find the corresponding y-coordinates using Equation 1 \( y_1 = x_1^3 \):
When \( x_1 = 0 \): \( y_1 = (0)^3 = 0 \). So, the point is \( (0, 0) \).
When \( x_1 = 3 \): \( y_1 = (3)^3 = 27 \). So, the point is \( (3, 27) \).
Therefore, the required points on the curve are \( (0, 0) \) and \( (3, 27) \).
In simple words: Find the curve's derivative. Set this derivative (the tangent's slope) equal to the point's \( y \)-coordinate from the original curve equation. Solve for \( x \), then find the matching \( y \) values to get the points.

Exam Tip: Carefully translate the problem statement into a mathematical equation. "Slope of the tangent is equal to the y-coordinate of the point" directly means \( \frac{dy}{dx} = y \).

Question 18. For the curve \( y = 4x^3 - 2x^5 \), find all the points at which the tangents pass through the origin.
Answer: The equation of the curve is \( y = 4x^3 - 2x^5 \).
Let \( (x_1, y_1) \) be a point of tangency on the curve. So, \( y_1 = 4x_1^3 - 2x_1^5 \). (Equation 1)
First, we find the derivative of \( y \) with respect to \( x \) to get the slope of the tangent:
\( \frac { dy }{ dx } = 12x^2 - 10x^4 \).
The slope of the tangent at \( (x_1, y_1) \) is \( m = 12x_1^2 - 10x_1^4 \).
The equation of the tangent line at \( (x_1, y_1) \) is given by \( y - y_1 = m(x - x_1) \).
\( y - y_1 = (12x_1^2 - 10x_1^4)(x - x_1) \).
Since the tangent passes through the origin \( (0, 0) \), we substitute \( x=0 \) and \( y=0 \) into the tangent equation:
\( 0 - y_1 = (12x_1^2 - 10x_1^4)(0 - x_1) \)
\( -y_1 = -x_1(12x_1^2 - 10x_1^4) \)
\( y_1 = x_1(12x_1^2 - 10x_1^4) \)
\( y_1 = 12x_1^3 - 10x_1^5 \). (Equation 2)
Now, we have two expressions for \( y_1 \). We equate Equation 1 and Equation 2:
\( 4x_1^3 - 2x_1^5 = 12x_1^3 - 10x_1^5 \).
Rearrange the terms to solve for \( x_1 \):
\( 12x_1^3 - 4x_1^3 - 10x_1^5 + 2x_1^5 = 0 \)
\( 8x_1^3 - 8x_1^5 = 0 \)
\( 8x_1^3(1 - x_1^2) = 0 \)
\( 8x_1^3(1 - x_1)(1 + x_1) = 0 \).
This equation gives three possible values for \( x_1 \):
\( x_1 = 0 \)
\( 1 - x_1 = 0 \implies x_1 = 1 \)
\( 1 + x_1 = 0 \implies x_1 = -1 \).
Now, we find the corresponding y-coordinates using Equation 1 \( y_1 = 4x_1^3 - 2x_1^5 \):
When \( x_1 = 0 \): \( y_1 = 4(0)^3 - 2(0)^5 = 0 \). Point: \( (0, 0) \).
When \( x_1 = 1 \): \( y_1 = 4(1)^3 - 2(1)^5 = 4 - 2 = 2 \). Point: \( (1, 2) \).
When \( x_1 = -1 \): \( y_1 = 4(-1)^3 - 2(-1)^5 = 4(-1) - 2(-1) = -4 + 2 = -2 \). Point: \( (-1, -2) \).
Therefore, the points on the curve where the tangents pass through the origin are \( (0, 0), (1, 2) \), and \( (-1, -2) \).
In simple words: Assume a point \( (x_1, y_1) \) on the curve. Find the tangent's slope at this point. Write the tangent's equation. Since the tangent passes through the origin, plug \( (0,0) \) into the tangent equation to find a relationship between \( x_1 \) and \( y_1 \). Solve this with the original curve equation to find all possible \( x_1 \) values, then find their matching \( y_1 \) values.

Exam Tip: When a tangent passes through a specific external point (like the origin), substitute that point's coordinates into the general equation of the tangent line. This generates a crucial relationship to solve for the point of tangency.

Question 19. Find the points on the curve \( x^2 + y^2 – 2x = 3 \) at which tangents are parallel to the x-axis.
Answer: The equation of the curve is \( x^2 + y^2 – 2x = 3 \). (Equation 1)
To find the slope of the tangent, we differentiate both sides with respect to \( x \):
\( 2x + 2y \frac { dy }{ dx } – 2 = 0 \).
Rearrange to solve for \( \frac{dy}{dx} \):
\( 2y \frac { dy }{ dx } = 2 - 2x \)
\( \frac { dy }{ dx } = \frac { 2 - 2x }{ 2y } = \frac { 1 - x }{ y } \).
If the tangent is parallel to the x-axis, its slope \( \frac{dy}{dx} \) must be \( 0 \).
Set \( \frac { 1 - x }{ y } = 0 \).
This implies that the numerator must be \( 0 \) (assuming \( y \neq 0 \)):
\( 1 - x = 0 \implies x = 1 \).
Now, substitute \( x = 1 \) into the original curve equation (Equation 1) to find the corresponding y-coordinates:
\( (1)^2 + y^2 – 2(1) = 3 \)
\( 1 + y^2 - 2 = 3 \)
\( y^2 - 1 = 3 \)
\( y^2 = 4 \)
\( y = \pm 2 \).
So, the points where the tangents are parallel to the x-axis are \( (1, 2) \) and \( (1, -2) \).
In simple words: Differentiate the curve's equation. Set the derivative (the slope of the tangent) to zero, as tangents parallel to the x-axis have no slope. Solve for \( x \). Use this \( x \) value in the original curve equation to find the corresponding \( y \) values, which will give you the required points.

Exam Tip: For implicit differentiation, remember to apply the chain rule to terms involving \( y \) (e.g., \( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \)). Also, horizontal tangents always have a slope of zero.

Question 20. Find the equation of the normal at the point \( (am^2, am^3) \) for the curve \( ay^2 = x^3 \).
Answer: The equation of the curve is \( ay^2 = x^3 \).
First, we differentiate both sides with respect to \( x \) to find \( \frac{dy}{dx} \):
\( a(2y) \frac { dy }{ dx } = 3x^2 \)
\( \frac { dy }{ dx } = \frac { 3x^2 }{ 2ay } \).
Now, we evaluate the slope of the tangent at the given point \( (am^2, am^3) \):
Slope of tangent \( m_t = \left(\frac { dy }{ dx }\right)_{(am^2, am^3)} = \frac { 3(am^2)^2 }{ 2a(am^3) } \)
\( m_t = \frac { 3a^2m^4 }{ 2a^2m^3 } = \frac { 3m }{ 2 } \).
The slope of the normal is the negative reciprocal of the tangent's slope:
Slope of normal \( m_n = - \frac{1}{m_t} = - \frac{1}{\frac{3m}{2}} = - \frac{2}{3m} \).
Now, we find the equation of the normal at the point \( (am^2, am^3) \) using the point-slope form \( y - y_1 = m_n(x - x_1) \):
\( y - am^3 = - \frac{2}{3m} (x - am^2) \).
Multiply both sides by \( 3m \) to clear the denominator:
\( 3m(y - am^3) = -2(x - am^2) \)
\( 3my - 3am^4 = -2x + 2am^2 \).
Rearrange the terms to get the equation in general form:
\( 2x + 3my - 3am^4 - 2am^2 = 0 \)
\( 2x + 3my - am^2(3m^2 + 2) = 0 \).
In simple words: Differentiate the curve's equation to find \( \frac{dy}{dx} \). Plug the given point into \( \frac{dy}{dx} \) to get the tangent's slope. Take the negative reciprocal for the normal's slope. Finally, use the given point and the normal's slope to write the normal's equation.

Exam Tip: Remember to use implicit differentiation correctly when \( y \) is not explicitly defined as a function of \( x \). Also, clearly distinguish between the slope of the tangent and the slope of the normal.

Question 21. Find the equation of the normal to the curve \( y = x^3 + 2x + 6 \) which is parallel to the line \( x + 14y + 4 = 0 \).
Answer: The equation of the curve is \( y = x^3 + 2x + 6 \). (Equation 1)
Let \( (x_1, y_1) \) be the point on the curve where the normal is drawn.
First, we find the derivative of \( y \) with respect to \( x \) to get the slope of the tangent:
\( \frac { dy }{ dx } = 3x^2 + 2 \).
The slope of the tangent at \( (x_1, y_1) \) is \( m_t = 3x_1^2 + 2 \).
The slope of the normal at \( (x_1, y_1) \) is \( m_n = - \frac{1}{3x_1^2 + 2} \).
The given line is \( x + 14y + 4 = 0 \). We find its slope by rearranging the equation:
\( 14y = -x - 4 \)
\( y = -\frac{1}{14}x - \frac{4}{14} \).
The slope of this line is \( -\frac{1}{14} \).
Since the normal is parallel to this line, their slopes must be equal:
\( - \frac{1}{3x_1^2 + 2} = -\frac{1}{14} \).
This implies \( 3x_1^2 + 2 = 14 \).
\( 3x_1^2 = 12 \)
\( x_1^2 = 4 \)
\( x_1 = \pm 2 \).
Now, we find the corresponding y-coordinates using Equation 1 \( y = x^3 + 2x + 6 \):
When \( x_1 = 2 \):
\( y_1 = (2)^3 + 2(2) + 6 = 8 + 4 + 6 = 18 \).
So, one point is \( (2, 18) \).
When \( x_1 = -2 \):
\( y_1 = (-2)^3 + 2(-2) + 6 = -8 - 4 + 6 = -6 \).
So, the other point is \( (-2, -6) \).
Now, we find the equation of the normal for each point, using the normal's slope \( m_n = -\frac{1}{14} \):
For point \( (2, 18) \):
\( y - 18 = -\frac{1}{14}(x - 2) \)
\( 14(y - 18) = -(x - 2) \)
\( 14y - 252 = -x + 2 \)
\( x + 14y - 254 = 0 \).
For point \( (-2, -6) \):
\( y - (-6) = -\frac{1}{14}(x - (-2)) \)
\( y + 6 = -\frac{1}{14}(x + 2) \)
\( 14(y + 6) = -(x + 2) \)
\( 14y + 84 = -x - 2 \)
\( x + 14y + 86 = 0 \).
Therefore, the equations of the normal are \( x + 14y - 254 = 0 \) and \( x + 14y + 86 = 0 \).
In simple words: Find the slope of the curve's normal. Find the slope of the given line. Since the normal is parallel to this line, their slopes are the same. Use this to find the \( x \) values on the curve. Then find the matching \( y \) values. Finally, write the normal's equation for each point using the point-slope form.

Exam Tip: Always clearly identify the slope of the tangent and the slope of the normal. If a line is parallel to another, their slopes are equal. If perpendicular, their slopes are negative reciprocals.

Question 22. Find the equations of the tangent and normal to the parabola \( y^2 = 4ax \) at \( (at^2, 2at) \).
Answer: The equation of the curve is \( y^2 = 4ax \).
First, we differentiate both sides implicitly with respect to \( x \):
\( 2y \frac { dy }{ dx } = 4a \)
\( \frac { dy }{ dx } = \frac { 4a }{ 2y } = \frac { 2a }{ y } \).
Now, we find the slope of the tangent at the given point \( (at^2, 2at) \):
Slope of tangent \( m_t = \left(\frac { dy }{ dx }\right)_{(at^2, 2at)} = \frac { 2a }{ 2at } = \frac { 1 }{ t } \).
Equation of the tangent at \( (at^2, 2at) \) using \( y - y_1 = m_t(x - x_1) \):
\( y - 2at = \frac { 1 }{ t } (x - at^2) \).
Multiply both sides by \( t \):
\( t(y - 2at) = x - at^2 \)
\( ty - 2at^2 = x - at^2 \)
\( ty = x + at^2 \).
So, the equation of the tangent is \( x - ty + at^2 = 0 \).
The slope of the normal is the negative reciprocal of the tangent's slope:
Slope of normal \( m_n = - \frac{1}{m_t} = - \frac{1}{\frac{1}{t}} = -t \).
Equation of the normal at \( (at^2, 2at) \) using \( y - y_1 = m_n(x - x_1) \):
\( y - 2at = -t(x - at^2) \)
\( y - 2at = -tx + at^3 \).
Rearrange the terms to get the equation in general form:
\( tx + y - 2at - at^3 = 0 \).
In simple words: Differentiate the parabola's equation to find \( \frac{dy}{dx} \). Plug the given parametric point into \( \frac{dy}{dx} \) to get the tangent's slope. Use this slope and the point to write the tangent's equation. For the normal, take the negative reciprocal of the tangent's slope, and use that new slope with the same point to write the normal's equation.

Exam Tip: For parametric points, ensure to use the x and y coordinates correctly when applying the point-slope form. Always simplify the final equations to their most compact form.

Question 23. Prove that the curves \(x = y^2\) and \(xy = k\) cut at right angles, if \(8k^2 = 1\).
Answer:
The two given curves are:
\(x = y^2\) ... (1)
and \(xy = k\) ... (2)
Substituting \(x = y^2\) into equation (2), we get
\(y^3 = k \implies y = k^{\frac{1}{3}}\).
From (1), \(x = y^2 = (k^{\frac{1}{3}})^2 = k^{\frac{2}{3}}\).
Therefore, the point where they intersect is \((k^{\frac{2}{3}}, k^{\frac{1}{3}})\).
Differentiating equation (1) with respect to \(x\), we get:
\(1 = 2y \frac{dy}{dx}\)
\( \implies \frac{dy}{dx} = \frac{1}{2y} \).
At \((k^{\frac{2}{3}}, k^{\frac{1}{3}})\), the slope is \(\frac{dy}{dx} = \frac{1}{2k^{\frac{1}{3}}}\).
Thus, the tangent's slope for the curve \(x = y^2\) at \((k^{\frac{2}{3}}, k^{\frac{1}{3}})\) is \(m_1 = \frac{1}{2k^{\frac{1}{3}}}\).
Differentiating \(xy = k\) with respect to \(x\), we get:
\(x \frac{dy}{dx} + y \cdot 1 = 0\)
\( \implies x \frac{dy}{dx} = -y \)
\( \implies \frac{dy}{dx} = -\frac{y}{x} \).
At \((k^{\frac{2}{3}}, k^{\frac{1}{3}})\), the slope is \(\frac{dy}{dx} = -\frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}} = -\frac{1}{k^{\frac{1}{3}}}\).
Consequently, the tangent's slope for the curve \(xy = k\) at \((k^{\frac{2}{3}}, k^{\frac{1}{3}})\) is \(m_2 = -\frac{1}{k^{\frac{1}{3}}}\).
The curves will intersect at right angles if the product of their tangent slopes at the intersection point is \(-1\).
So, \(m_1 \cdot m_2 = -1\).
\((\frac{1}{2k^{\frac{1}{3}}})(-\frac{1}{k^{\frac{1}{3}}}) = -1\)
\( \implies -\frac{1}{2k^{\frac{2}{3}}} = -1 \)
\( \implies 1 = 2k^{\frac{2}{3}} \)
Cubing both sides:
\( \implies 1^3 = (2k^{\frac{2}{3}})^3 \)
\( \implies 1 = 8k^2 \).
Hence, the statement is proved.
In simple words: To prove curves meet at right angles, find the slope of the tangent for each curve at their meeting point. If you multiply these two slopes and the result is -1, then the curves cross at a 90-degree angle.

Exam Tip: Remember that two curves intersect at right angles if the product of their tangent slopes at the intersection point is -1. Ensure all differentiation steps are correct and algebraic manipulations are accurate.

Question 24. Find the equations of the tangent and normal to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) at the point \((x_0, y_0)\).
Answer:
Given the hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).
By differentiating with respect to \(x\), we obtain:
\(\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0\)
\( \implies \frac{2y}{b^2} \frac{dy}{dx} = \frac{2x}{a^2} \)
\( \implies \frac{dy}{dx} = \frac{2x}{a^2} \cdot \frac{b^2}{2y} = \frac{b^2x}{a^2y} \).
Now, at the point \((x_0, y_0)\), the derivative is:
\(\frac{dy}{dx}_{(x_0, y_0)} = \frac{b^2x_0}{a^2y_0}\).
Therefore, the slope of the tangent \(m = \frac{b^2x_0}{a^2y_0}\).
Next, the equation for the tangent at \((x_0, y_0)\) is:
\(y - y_0 = m(x - x_0)\)
\(y - y_0 = \frac{b^2x_0}{a^2y_0} (x - x_0)\)
\(a^2y_0(y - y_0) = b^2x_0(x - x_0)\)
\(a^2yy_0 - a^2y_0^2 = b^2xx_0 - b^2x_0^2\)
\(b^2xx_0 - a^2yy_0 = b^2x_0^2 - a^2y_0^2\).
Divide this entire equation by \(a^2b^2\):
\(\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2}\).
As the point \((x_0, y_0)\) is on the curve \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), it means \(\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1\).
So, the equation of the tangent becomes:
\( \implies \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1 \). (Equation of Tangent)
After that, the slope of the normal \(m_n = -\frac{1}{m} = -\frac{1}{\frac{b^2x_0}{a^2y_0}} = -\frac{a^2y_0}{b^2x_0}\).
Therefore, the equation of the normal is:
\(y - y_0 = m_n(x - x_0)\)
\(y - y_0 = -\frac{a^2y_0}{b^2x_0} (x - x_0)\)
\(b^2x_0(y - y_0) = -a^2y_0(x - x_0)\)
\(a^2y_0(x - x_0) + b^2x_0(y - y_0) = 0\).
Dividing by \(a^2b^2x_0y_0\) gives an alternative form:
\(\frac{x-x_0}{b^2x_0} + \frac{y-y_0}{a^2y_0} = 0\).
In simple words: First, find how fast the hyperbola's \(y\) changes with \(x\) (its derivative). This gives you the tangent's slope. Use this slope with the point \((x_0, y_0)\) to write the tangent's equation. For the normal, its slope is the negative inverse of the tangent's slope. Then use that new slope with \((x_0, y_0)\) to find the normal's equation.

Exam Tip: Remember the standard formula for the equation of a tangent \(y - y_1 = m(x - x_1)\) and normal \(y - y_1 = -\frac{1}{m}(x - x_1)\) to a curve at a given point. Be careful with algebraic rearrangements to match the required forms.

Question 25. Find the equation of the tangent to the curve \(y = \sqrt{3x-2}\), which is parallel to the line \(4x - 2y + 5 = 0\).
Answer:
We assume the tangent line's contact point is \(P(x_1, y_1)\).
The equation of the curve is \(y = \sqrt{3x-2}\).
By differentiating both sides concerning \(x\), we obtain:
\(\frac{dy}{dx} = \frac{1}{2\sqrt{3x-2}} \cdot 3 = \frac{3}{2\sqrt{3x-2}}\).
At \((x_1, y_1)\), the slope of the tangent is \(\frac{dy}{dx}_{(x_1, y_1)} = \frac{3}{2\sqrt{3x_1-2}}\).
As the tangent at the point \((x_1, y_1)\) runs parallel to the given line \(4x - 2y + 5 = 0\), their slopes must be equal.
The slope of the line \(4x - 2y + 5 = 0\) is found by rearranging it into \(y = mx + c\) form:
\(2y = 4x + 5 \implies y = 2x + \frac{5}{2}\).
So, its slope is \(m_{line} = 2\).
Consequently, we equate the slopes:
\(\frac{3}{2\sqrt{3x_1-2}} = 2\)
\(3 = 4\sqrt{3x_1-2}\).
Square both sides to get:
\(3^2 = (4\sqrt{3x_1-2})^2\)
\(9 = 16(3x_1-2)\)
\(9 = 48x_1 - 32\)
\(48x_1 = 41\)
\(x_1 = \frac{41}{48}\).
Given that point \((x_1, y_1)\) is on the curve \(y = \sqrt{3x-2}\), we can find \(y_1\):
\(y_1 = \sqrt{3x_1-2} = \sqrt{3(\frac{41}{48})-2}\)
\(y_1 = \sqrt{\frac{41}{16}-2} = \sqrt{\frac{41-32}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}\).
Thus, the point where contact occurs is \((\frac{41}{48}, \frac{3}{4})\).
Therefore, the equation for the desired tangent line is:
\(y - y_1 = m(x - x_1)\)
\(y - \frac{3}{4} = 2(x - \frac{41}{48})\).
To simplify, multiply the entire equation by the LCM of 4 and 48, which is 48:
\(48(y - \frac{3}{4}) = 48 \cdot 2(x - \frac{41}{48})\)
\(48y - 36 = 96(x - \frac{41}{48})\)
\(48y - 36 = 96x - 41 \cdot 2\)
\(48y - 36 = 96x - 82\)
\(96x - 48y = 82 - 36\)
\(96x - 48y = 46\).
Divide by 2:
\(48x - 24y = 23\).
In simple words: First, find the slope of the curve by taking its derivative. Then, find the slope of the line it needs to be parallel to. Since they are parallel, set these two slopes equal to each other to find the \(x\)-coordinate. Use the \(x\)-coordinate in the original curve equation to find the \(y\)-coordinate. Finally, use the point and the slope to write the equation of the tangent line.

Exam Tip: For parallel lines, slopes are equal. For perpendicular lines, the product of slopes is -1. Remember to square both sides carefully when solving for a variable inside a square root and to simplify the final equation.

Question 27. The line \(y = x + 1\) is a tangent to the curve \(y^2 = 4x\) at the point
(a) (1, 2)
(b) (2, 1)
(c) (1, - 2)
(d) (- 1, 2)
Answer: (a) (1, 2)
The given curve is \(y^2 = 4x\).
By differentiating implicitly with respect to \(x\), we find \(2y \frac{dy}{dx} = 4\).
\( \implies \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \).
The line \(y = x + 1\) has a slope of 1.
Since the line is a tangent to the curve, their slopes must be equal at the point of contact.
So, \(\frac{2}{y} = 1 \implies y = 2\).
Substituting \(y = 2\) into the curve's equation \(y^2 = 4x\), we obtain:
\(2^2 = 4x \implies 4 = 4x \implies x = 1\).
The point of contact is \((1, 2)\).
Thus, the correct option is (a).
In simple words: First, find the slope of the curve using differentiation. Then, find the slope of the given line. Since the line is a tangent, their slopes must be equal at the point of contact. Use this to find the \(y\)-coordinate, and then the \(x\)-coordinate from the curve's equation.

Exam Tip: For tangency problems, always equate the derivative of the curve (which gives the tangent's slope) to the slope of the given tangent line. Then, use the curve's equation to find the corresponding coordinates.