OP Malhotra Class 12 Maths Solutions Chapter 11 Applications of Derivatives Exercise 11 (A)

Question 1. Find the slopes of the following curves:
(i) \( y = \frac { 4 }{ x } \) at the point (2, 2)
(ii) \( y = 2x^2 - 1 \) at \( x = 1 \)
(iii) \( y = 2x - x^2 \) at \( x = 1 \)
(iv) \( y = x^2 - \sin x \) at \( x = 0 \)
Answer:
(i) Given curve is \( y = \frac{4}{x} \).
Differentiating both sides with respect to \( x \), we get:
\( \frac{dy}{dx} = -\frac{4}{x^2} \)
Now, we find the slope of the tangent at the point (2, 2):
\( \left(\frac{dy}{dx}\right)_{(2,2)} = \frac{-4}{2^2} = \frac{-4}{4} = -1 \)

(ii) Given curve is \( y = 2x^2 - 1 \).
Differentiating both sides with respect to \( x \), we get:
\( \frac{dy}{dx} = 4x \)
Now, we find the slope of the tangent at \( x = 1 \):
\( \left(\frac{dy}{dx}\right)_{x=1} = 4 \times 1 = 4 \)

(iii) Given curve is \( y = 2x - x^2 \).
Differentiating both sides with respect to \( x \), we get:
\( \frac{dy}{dx} = 2 - 2x \)
Now, we find the slope of the tangent at \( x = 1 \):
\( \left(\frac{dy}{dx}\right)_{x=1} = 2 - 2(1) = 2 - 2 = 0 \)

(iv) Given curve is \( y = x^2 - \sin x \).
Differentiating both sides with respect to \( x \), we get:
\( \frac{dy}{dx} = 2x - \cos x \)
Now, we find the slope of the tangent at \( x = 0 \):
\( \left(\frac{dy}{dx}\right)_{x=0} = 2(0) - \cos 0 = 0 - 1 = -1 \)
In simple words: To find the slope of a curve at a certain point, first you need to find the derivative of the curve's equation. Then, you plug in the x-coordinate of the point into this derivative. The result is the slope of the line that just touches the curve at that exact point.

🎯 Exam Tip: Remember that the derivative \( \frac{dy}{dx} \) gives the slope of the tangent line to the curve at any point \( (x, y) \). For specific points, always substitute the coordinates after finding the derivative.

Question 2. Find the slope of the normal to the curve \( y = 3x^2 \) at the point whose x-coordinate is 2.
Answer:
The given equation of the curve is \( y = 3x^2 \).
When \( x = 2 \), we find the corresponding y-coordinate:
\( y = 3(2)^2 = 3 \times 4 = 12 \)
So, the point on the curve is (2, 12).
Now, we differentiate the equation of the curve with respect to \( x \) to find the slope of the tangent:
\( \frac{dy}{dx} = 6x \)
At the point (2, 12), the slope of the tangent is:
\( \left(\frac{dy}{dx}\right)_{(2,12)} = 6 \times 2 = 12 \)
The slope of the normal to the curve at a point is the negative reciprocal of the slope of the tangent at that same point. So, the slope of the normal is:
\( \text{Slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{12} \)
In simple words: First, find the y-value for the given x-value to get the full point. Then, calculate the derivative of the curve's equation and put the x-value into it to find the slope of the tangent. The slope of the normal line is simply the flipped and negative version of the tangent's slope.

🎯 Exam Tip: Always remember that the slope of the normal is \( -\frac{1}{m_{\text{tangent}}} \). A common mistake is to forget the negative sign or to use the reciprocal without changing the sign.

Question 3. Find the equation of the tangent and normal to the given curves at the points given:
(i) \( y = 2x^2 - 3x - 1 \) at the point (1, 2)
(ii) \( x = \cos t, y = \sin t, \) at \( t = \frac{\pi}{4} \)
(iii) \( y = x^2 + 4x + 1 \) at the point whose abscissa is 3
(iv) \( y^2 = \frac{x^3}{4-x} \) at (2, -2)
(v) \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at \( (x_0, y_0) \)
Answer:
(i) Given equation of curve is \( y = 2x^2 - 3x - 1 \).
Differentiate with respect to \( x \):
\( \frac{dy}{dx} = 4x - 3 \)
At the point (1, 2), the slope of the tangent is:
\( m_{\text{tangent}} = \left(\frac{dy}{dx}\right)_{(1,2)} = 4(1) - 3 = 1 \)
The equation of the tangent at (1, 2) is \( y - y_1 = m_{\text{tangent}}(x - x_1) \):
\( y - 2 = 1(x - 1) \)
\( y - 2 = x - 1 \)
\( \implies x - y + 1 = 0 \)
The slope of the normal is \( m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{1} = -1 \).
The equation of the normal at (1, 2) is \( y - y_1 = m_{\text{normal}}(x - x_1) \):
\( y - 2 = -1(x - 1) \)
\( y - 2 = -x + 1 \)
\( \implies x + y - 3 = 0 \)

(ii) Given parametric equations of the curve are \( x = \cos t \) and \( y = \sin t \).
Differentiate both equations with respect to \( t \):
\( \frac{dx}{dt} = -\sin t \)
\( \frac{dy}{dt} = \cos t \)
The slope of the tangent \( \frac{dy}{dx} \) is:
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t}{-\sin t} = -\cot t \)
At \( t = \frac{\pi}{4} \):
\( x = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
\( y = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
So the point is \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \).
The slope of the tangent is:
\( m_{\text{tangent}} = -\cot \frac{\pi}{4} = -1 \)
The equation of the tangent at \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) is \( y - y_1 = m_{\text{tangent}}(x - x_1) \):
\( y - \frac{1}{\sqrt{2}} = -1\left(x - \frac{1}{\sqrt{2}}\right) \)
\( y - \frac{1}{\sqrt{2}} = -x + \frac{1}{\sqrt{2}} \)
\( \implies x + y - \frac{2}{\sqrt{2}} = 0 \)
\( \implies x + y - \sqrt{2} = 0 \)
The slope of the normal is \( m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-1} = 1 \).
The equation of the normal at \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) is \( y - y_1 = m_{\text{normal}}(x - x_1) \):
\( y - \frac{1}{\sqrt{2}} = 1\left(x - \frac{1}{\sqrt{2}}\right) \)
\( y - \frac{1}{\sqrt{2}} = x - \frac{1}{\sqrt{2}} \)
\( \implies y = x \)

(iii) Given equation of curve is \( y = x^2 + 4x + 1 \).
The abscissa (x-coordinate) is \( x = 3 \). Find the y-coordinate:
\( y = (3)^2 + 4(3) + 1 = 9 + 12 + 1 = 22 \)
So the point of contact is (3, 22).
Differentiate the curve with respect to \( x \):
\( \frac{dy}{dx} = 2x + 4 \)
At the point (3, 22), the slope of the tangent is:
\( m_{\text{tangent}} = \left(\frac{dy}{dx}\right)_{(3,22)} = 2(3) + 4 = 6 + 4 = 10 \)
The equation of the tangent at (3, 22) is \( y - y_1 = m_{\text{tangent}}(x - x_1) \):
\( y - 22 = 10(x - 3) \)
\( y - 22 = 10x - 30 \)
\( \implies 10x - y - 8 = 0 \)
The slope of the normal is \( m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{10} \).
The equation of the normal at (3, 22) is \( y - y_1 = m_{\text{normal}}(x - x_1) \):
\( y - 22 = -\frac{1}{10}(x - 3) \)
\( 10(y - 22) = -(x - 3) \)
\( 10y - 220 = -x + 3 \)
\( \implies x + 10y - 223 = 0 \)

(iv) Given equation of curve is \( y^2 = \frac{x^3}{4-x} \).
Differentiate implicitly with respect to \( x \) using the quotient rule:
\( 2y \frac{dy}{dx} = \frac{(4-x)(3x^2) - x^3(-1)}{(4-x)^2} \)
\( 2y \frac{dy}{dx} = \frac{12x^2 - 3x^3 + x^3}{(4-x)^2} \)
\( 2y \frac{dy}{dx} = \frac{12x^2 - 2x^3}{(4-x)^2} \)
\( \frac{dy}{dx} = \frac{12x^2 - 2x^3}{2y(4-x)^2} = \frac{6x^2 - x^3}{y(4-x)^2} \)
At the point (2, -2), the slope of the tangent is:
\( m_{\text{tangent}} = \left(\frac{dy}{dx}\right)_{(2,-2)} = \frac{6(2)^2 - (2)^3}{(-2)(4-2)^2} \)
\( = \frac{6(4) - 8}{(-2)(2)^2} = \frac{24 - 8}{(-2)(4)} = \frac{16}{-8} = -2 \)
The equation of the tangent at (2, -2) is \( y - y_1 = m_{\text{tangent}}(x - x_1) \):
\( y - (-2) = -2(x - 2) \)
\( y + 2 = -2x + 4 \)
\( \implies 2x + y - 2 = 0 \)
The slope of the normal is \( m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-2} = \frac{1}{2} \).
The equation of the normal at (2, -2) is \( y - y_1 = m_{\text{normal}}(x - x_1) \):
\( y - (-2) = \frac{1}{2}(x - 2) \)
\( 2(y + 2) = x - 2 \)
\( 2y + 4 = x - 2 \)
\( \implies x - 2y - 6 = 0 \)

(v) Given equation of curve is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Differentiate implicitly with respect to \( x \):
\( \frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \)
\( \frac{2y}{b^2} \frac{dy}{dx} = \frac{2x}{a^2} \)
\( \frac{dy}{dx} = \frac{2x}{a^2} \times \frac{b^2}{2y} = \frac{b^2x}{a^2y} \)
At the point \( (x_0, y_0) \), the slope of the tangent is:
\( m_{\text{tangent}} = \left(\frac{dy}{dx}\right)_{(x_0,y_0)} = \frac{b^2x_0}{a^2y_0} \)
The equation of the tangent at \( (x_0, y_0) \) is \( y - y_0 = m_{\text{tangent}}(x - x_0) \):
\( y - y_0 = \frac{b^2x_0}{a^2y_0}(x - x_0) \)
Multiply by \( a^2y_0 \):
\( a^2y_0(y - y_0) = b^2x_0(x - x_0) \)
\( a^2yy_0 - a^2y_0^2 = b^2xx_0 - b^2x_0^2 \)
Rearrange terms:
\( b^2xx_0 - a^2yy_0 = b^2x_0^2 - a^2y_0^2 \)
Divide by \( a^2b^2 \):
\( \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \)
Since \( (x_0, y_0) \) lies on the curve \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we know that \( \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1 \).
So, the equation of the tangent is:
\( \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1 \)
The slope of the normal is \( m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{a^2y_0}{b^2x_0} \).
The equation of the normal at \( (x_0, y_0) \) is \( y - y_0 = m_{\text{normal}}(x - x_0) \):
\( y - y_0 = -\frac{a^2y_0}{b^2x_0}(x - x_0) \)
Multiply by \( b^2x_0 \):
\( b^2x_0(y - y_0) = -a^2y_0(x - x_0) \)
\( b^2x_0y - b^2x_0y_0 = -a^2y_0x + a^2x_0y_0 \)
\( a^2y_0x + b^2x_0y = a^2x_0y_0 + b^2x_0y_0 \)
\( a^2y_0x + b^2x_0y = (a^2 + b^2)x_0y_0 \)
Divide by \( x_0y_0 \):
\( \frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2 \)
In simple words: For each part, first find the slope of the tangent by differentiating the curve's equation and plugging in the point's coordinates. Use this slope and point to write the tangent's equation. For the normal, the slope is the negative inverse of the tangent's slope, and then you use this new slope with the same point to write the normal's equation. Always simplify your equations to a standard form.

🎯 Exam Tip: When dealing with parametric equations, remember to use the chain rule to find \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). For implicit differentiation, differentiate each term with respect to x, remembering to multiply \( \frac{dy}{dx} \) for terms involving \( y \).

Question 4. Find the equation of the tangent and normal to the parabola \( y^2 = 4ax \) at \( (at^2, 2at) \).
Answer:
The given equation of the parabola is \( y^2 = 4ax \).
Differentiate both sides of the equation with respect to \( x \):
\( 2y \frac{dy}{dx} = 4a \)
\( \implies \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \)
At the point \( (at^2, 2at) \), the slope of the tangent is:
\( m_{\text{tangent}} = \left(\frac{dy}{dx}\right)_{(at^2,2at)} = \frac{2a}{2at} = \frac{1}{t} \)
The equation of the tangent at \( (at^2, 2at) \) is \( y - y_1 = m_{\text{tangent}}(x - x_1) \):
\( y - 2at = \frac{1}{t}(x - at^2) \)
Multiply by \( t \):
\( ty - 2at^2 = x - at^2 \)
\( \implies x - ty + at^2 = 0 \)
The slope of the normal is \( m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{1/t} = -t \).
The equation of the normal at \( (at^2, 2at) \) is \( y - y_1 = m_{\text{normal}}(x - x_1) \):
\( y - 2at = -t(x - at^2) \)
\( y - 2at = -tx + at^3 \)
\( \implies tx + y - 2at - at^3 = 0 \)
In simple words: To find the tangent and normal equations for a parabola at a specific point, first differentiate the parabola's equation to get the slope of the tangent. Use this slope and the given point to form the tangent equation. For the normal line, use the negative reciprocal of the tangent's slope with the same point to form its equation.

🎯 Exam Tip: When working with general points like \( (at^2, 2at) \), keep the variable \( t \) in your expressions. The process is the same as with numeric points: find the derivative, substitute coordinates for the slope, then use the point-slope formula for the equations.

Question 5. Find the equations to the tangent to the curve \( y = (x^2 - 1) (x - 2) \) at the points where the curve cuts the x-axis.
Answer:
The given equation of the curve is \( y = (x^2 - 1) (x - 2) \).
The curve cuts the x-axis when \( y = 0 \). So, we set \( y = 0 \):
\( (x^2 - 1) (x - 2) = 0 \)
This gives us \( x^2 - 1 = 0 \) or \( x - 2 = 0 \).
\( x^2 = 1 \implies x = \pm 1 \)
\( x = 2 \)
So, the points where the curve cuts the x-axis are (2, 0), (1, 0), and (-1, 0).
First, expand the equation of the curve:
\( y = x^3 - 2x^2 - x + 2 \)
Now, differentiate with respect to \( x \) to find the slope of the tangent:
\( \frac{dy}{dx} = 3x^2 - 4x - 1 \)

**At point (2, 0):**
The slope of the tangent is \( m_{\text{tangent}} = \left(\frac{dy}{dx}\right)_{(2,0)} = 3(2)^2 - 4(2) - 1 = 3(4) - 8 - 1 = 12 - 8 - 1 = 3 \).
The equation of the tangent is \( y - y_1 = m_{\text{tangent}}(x - x_1) \):
\( y - 0 = 3(x - 2) \)
\( \implies 3x - y - 6 = 0 \)

**At point (1, 0):**
The slope of the tangent is \( m_{\text{tangent}} = \left(\frac{dy}{dx}\right)_{(1,0)} = 3(1)^2 - 4(1) - 1 = 3 - 4 - 1 = -2 \).
The equation of the tangent is \( y - y_1 = m_{\text{tangent}}(x - x_1) \):
\( y - 0 = -2(x - 1) \)
\( y = -2x + 2 \)
\( \implies 2x + y - 2 = 0 \)

**At point (-1, 0):**
The slope of the tangent is \( m_{\text{tangent}} = \left(\frac{dy}{dx}\right)_{(-1,0)} = 3(-1)^2 - 4(-1) - 1 = 3 + 4 - 1 = 6 \).
The equation of the tangent is \( y - y_1 = m_{\text{tangent}}(x - x_1) \):
\( y - 0 = 6(x + 1) \)
\( y = 6x + 6 \)
\( \implies 6x - y + 6 = 0 \)
In simple words: First, find all the points where the curve crosses the x-axis by setting y to zero. Next, differentiate the curve's equation to find the formula for the slope of the tangent line. For each of the points you found, plug its x-value into the slope formula. Finally, use each point and its unique slope to write the equation for each tangent line.

🎯 Exam Tip: Always factorize the curve equation to easily find the x-intercepts. Be careful with calculations for each point; a small error can propagate through the tangent equation.

Question 6. Find the point on the curve \( y = 2x^2 - 6x - 4 \) at which the tangent is parallel to the x-axis.
Answer:
The given equation of the curve is \( y = 2x^2 - 6x - 4 \).
When the tangent to a curve is parallel to the x-axis, its slope is 0.
First, differentiate the curve's equation with respect to \( x \) to find the slope of the tangent:
\( \frac{dy}{dx} = 4x - 6 \)
Set the slope to 0:
\( 4x - 6 = 0 \)
\( 4x = 6 \)
\( x = \frac{6}{4} = \frac{3}{2} \)
Now, substitute \( x = \frac{3}{2} \) back into the original curve equation to find the corresponding y-coordinate:
\( y = 2\left(\frac{3}{2}\right)^2 - 6\left(\frac{3}{2}\right) - 4 \)
\( y = 2\left(\frac{9}{4}\right) - 9 - 4 \)
\( y = \frac{9}{2} - 13 \)
\( y = \frac{9 - 26}{2} = -\frac{17}{2} \)
So, the required point on the curve is \( \left(\frac{3}{2}, -\frac{17}{2}\right) \).
In simple words: If a tangent line is flat (parallel to the x-axis), its slope is zero. So, you find the derivative of the curve's equation and set it equal to zero. This will give you the x-coordinate. Then, put this x-coordinate back into the original curve equation to find the matching y-coordinate. This gives you the full point.

🎯 Exam Tip: Remember that a tangent parallel to the x-axis implies \( \frac{dy}{dx} = 0 \). This is a key condition for finding local maxima or minima of a function.

Question 7. Find the points on the curve \( y = 12x - x^3 \) at which the slope is zero.
Answer:
The given equation of the curve is \( y = 12x - x^3 \).
We are looking for points where the slope of the tangent is zero.
First, differentiate the curve's equation with respect to \( x \):
\( \frac{dy}{dx} = 12 - 3x^2 \)
Set the slope to zero:
\( 12 - 3x^2 = 0 \)
\( 3x^2 = 12 \)
\( x^2 = 4 \)
\( x = \pm 2 \)
Now, we find the corresponding y-coordinates for these x-values:
When \( x = 2 \):
\( y = 12(2) - (2)^3 = 24 - 8 = 16 \)
So, one point is (2, 16).
When \( x = -2 \):
\( y = 12(-2) - (-2)^3 = -24 - (-8) = -24 + 8 = -16 \)
So, the other point is (-2, -16).
Thus, the required points on the curve where the slope is zero are (2, 16) and (-2, -16).
In simple words: To find where the curve has a flat slope (zero slope), you first take the derivative of the curve's formula. Then, set this derivative equal to zero and solve for x. For each x-value you find, plug it back into the curve's original equation to get the y-value. These (x, y) pairs are the points where the slope is zero.

🎯 Exam Tip: Remember to find both positive and negative roots when solving \( x^2 = \text{constant} \). Always substitute both x-values back into the original curve equation, not the derivative, to find the correct y-coordinates.

Question 8. Find the points on the curve \( x^2 + y^2 = 25 \), the tangents at which are (i) parallel to the x-axis, (ii) parallel to the y-axis.
Answer:
The given equation of the curve is \( x^2 + y^2 = 25 \). This is a circle centered at the origin with a radius of 5.
Differentiate the equation implicitly with respect to \( x \):
\( 2x + 2y \frac{dy}{dx} = 0 \)
\( 2y \frac{dy}{dx} = -2x \)
\( \implies \frac{dy}{dx} = -\frac{x}{y} \)

(i) **Tangent parallel to the x-axis:**
If the tangent is parallel to the x-axis, its slope \( \frac{dy}{dx} \) must be 0.
So, \( -\frac{x}{y} = 0 \)
This means \( x = 0 \).
Substitute \( x = 0 \) back into the original equation of the circle:
\( (0)^2 + y^2 = 25 \)
\( y^2 = 25 \)
\( y = \pm 5 \)
The points where the tangent is parallel to the x-axis are (0, 5) and (0, -5).

(ii) **Tangent parallel to the y-axis:**
If the tangent is parallel to the y-axis, its slope \( \frac{dy}{dx} \) is undefined. This happens when the denominator of \( \frac{dy}{dx} \) is 0. So, \( y = 0 \).
Alternatively, we can find \( \frac{dx}{dy} \) and set it to 0. Differentiating \( x^2 + y^2 = 25 \) with respect to \( y \):
\( 2x \frac{dx}{dy} + 2y = 0 \)
\( 2x \frac{dx}{dy} = -2y \)
\( \implies \frac{dx}{dy} = -\frac{y}{x} \)
Set \( \frac{dx}{dy} = 0 \):
\( -\frac{y}{x} = 0 \)
This means \( y = 0 \).
Substitute \( y = 0 \) back into the original equation of the circle:
\( x^2 + (0)^2 = 25 \)
\( x^2 = 25 \)
\( x = \pm 5 \)
The points where the tangent is parallel to the y-axis are (5, 0) and (-5, 0).
In simple words: For a circle, first find the derivative using implicit differentiation. If the tangent is flat (parallel to the x-axis), the slope is zero, so set the numerator of the derivative to zero and solve for x. If the tangent is straight up or down (parallel to the y-axis), the slope is undefined, so set the denominator of the derivative to zero and solve for y. Then use these x or y values to find the full coordinates from the original circle equation.

🎯 Exam Tip: For curves like circles, tangents parallel to the x-axis occur where \( \frac{dy}{dx} = 0 \), and tangents parallel to the y-axis occur where \( \frac{dx}{dy} = 0 \) (or \( \frac{dy}{dx} \) is undefined). Geometrically, these are the topmost/bottommost and leftmost/rightmost points of the circle, respectively.

Question 9. Find the point on the curve \( y^2 = 4x \), the tangent at which is parallel to the straight line \( y = 2x + 4 \).
Answer:
The given equation of the curve is \( y^2 = 4x \).
The given straight line is \( y = 2x + 4 \). The slope of this line is \( m_{\text{line}} = 2 \).
Since the tangent to the curve is parallel to this line, their slopes must be equal. So, the slope of the tangent to the curve is also 2.
First, differentiate the curve's equation implicitly with respect to \( x \):
\( 2y \frac{dy}{dx} = 4 \)
\( \implies \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \)
Set the slope of the tangent equal to the slope of the given line:
\( \frac{2}{y} = 2 \)
\( 2 = 2y \)
\( \implies y = 1 \)
Now, substitute \( y = 1 \) back into the original curve equation \( y^2 = 4x \) to find the x-coordinate:
\( (1)^2 = 4x \)
\( 1 = 4x \)
\( \implies x = \frac{1}{4} \)
Thus, the required point on the curve is \( \left(\frac{1}{4}, 1\right) \).
In simple words: First, find the slope of the given straight line. Since the curve's tangent is parallel to this line, the tangent's slope must be the same. Next, find the derivative of the curve's equation. Set this derivative equal to the slope you just found. Solve for the y-coordinate. Finally, plug this y-coordinate back into the curve's original equation to find the x-coordinate. This gives you the full point.

🎯 Exam Tip: The condition "parallel to a straight line" means their slopes are equal. Always find the slope of the given line first (by writing it in \( y=mx+c \) form if needed), and then set it equal to the derivative of the curve.

Question 10. Find the equation of the tangent line to \( y = 2x^2 + 1 \) which is parallel to the line \( 4x - y + 3 = 0 \).
Answer:
The given equation of the curve is \( y = 2x^2 + 7 \). (Note: The solution is based on \( y = 2x^2 + 7 \) as per the provided steps.)
The given straight line is \( 4x - y + 3 = 0 \). We can rewrite this in slope-intercept form (\( y = mx + c \)):
\( y = 4x + 3 \)
The slope of this line is \( m_{\text{line}} = 4 \).
Since the tangent to the curve is parallel to this line, the slope of the tangent must also be 4.
First, differentiate the curve's equation with respect to \( x \):
\( \frac{dy}{dx} = 4x \)
Set the slope of the tangent equal to 4:
\( 4x = 4 \)
\( \implies x = 1 \)
Now, substitute \( x = 1 \) back into the curve equation \( y = 2x^2 + 7 \) to find the y-coordinate of the point of contact:
\( y = 2(1)^2 + 7 = 2(1) + 7 = 2 + 7 = 9 \)
So, the point of contact (the point on the curve where the tangent touches) is (1, 9).
Now, we find the equation of the tangent line using the point (1, 9) and the slope \( m_{\text{tangent}} = 4 \):
\( y - y_1 = m_{\text{tangent}}(x - x_1) \)
\( y - 9 = 4(x - 1) \)
\( y - 9 = 4x - 4 \)
\( \implies 4x - y + 5 = 0 \)
This is the required equation of the tangent line.
In simple words: First, find the slope of the straight line given. Since the tangent line is parallel, it will have the same slope. Next, find the derivative of the curve's equation and set it equal to this slope to find the x-coordinate where the tangent touches. Use this x-coordinate in the original curve equation to find the y-coordinate. Finally, use this point and the slope to write the equation of the tangent line.

🎯 Exam Tip: Carefully identify the slope of the given line, as this will be the slope of your tangent. Always find the complete point of tangency (x and y coordinates) before writing the equation of the tangent line.

Question 11. Find the equation of the normal line to \( y = x^3 + 2x + 6 \) which is parallel to the line \( 14y + x + 4 = 0 \).
Answer:
The given equation of the curve is \( y = x^3 + 2x + 6 \).
The given straight line is \( 14y + x + 4 = 0 \). We rewrite this in slope-intercept form:
\( 14y = -x - 4 \)
\( y = -\frac{1}{14}x - \frac{4}{14} \)
The slope of this line is \( m_{\text{line}} = -\frac{1}{14} \).
Since the normal to the curve is parallel to this line, the slope of the normal must also be \( -\frac{1}{14} \).
First, differentiate the curve's equation with respect to \( x \) to find the slope of the tangent:
\( \frac{dy}{dx} = 3x^2 + 2 \)
The slope of the normal is the negative reciprocal of the slope of the tangent:
\( m_{\text{normal}} = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{3x^2 + 2} \)
Set the slope of the normal equal to the slope of the given line:
\( -\frac{1}{3x^2 + 2} = -\frac{1}{14} \)
\( \implies 3x^2 + 2 = 14 \)
\( 3x^2 = 12 \)
\( x^2 = 4 \)
\( x = \pm 2 \)
Now, find the corresponding y-coordinates for these x-values from the curve equation \( y = x^3 + 2x + 6 \):
When \( x = 2 \):
\( y = (2)^3 + 2(2) + 6 = 8 + 4 + 6 = 18 \)
So, one point of contact is (2, 18).
When \( x = -2 \):
\( y = (-2)^3 + 2(-2) + 6 = -8 - 4 + 6 = -6 \)
So, the other point of contact is (-2, -6).

Now, we find the equation of the normal line for each point using \( m_{\text{normal}} = -\frac{1}{14} \):
**At point (2, 18):**
\( y - y_1 = m_{\text{normal}}(x - x_1) \)
\( y - 18 = -\frac{1}{14}(x - 2) \)
\( 14(y - 18) = -(x - 2) \)
\( 14y - 252 = -x + 2 \)
\( \implies x + 14y - 254 = 0 \)

**At point (-2, -6):**
\( y - y_1 = m_{\text{normal}}(x - x_1) \)
\( y - (-6) = -\frac{1}{14}(x - (-2)) \)
\( y + 6 = -\frac{1}{14}(x + 2) \)
\( 14(y + 6) = -(x + 2) \)
\( 14y + 84 = -x - 2 \)
\( \implies x + 14y + 86 = 0 \)
In simple words: First, find the slope of the given straight line. Since the curve's normal line is parallel to it, the normal's slope is the same. Then, find the derivative of the curve for the tangent's slope, and from that, find the normal's slope formula. Set the normal's slope formula equal to the line's slope to find the x-coordinates. Plug these x-coordinates into the original curve equation to get the y-coordinates. Finally, use each point and the normal's slope to write the equation for each normal line.

🎯 Exam Tip: Distinguish carefully between the slope of the tangent \( (\frac{dy}{dx}) \) and the slope of the normal \( (-\frac{1}{dy/dx}) \). Ensure you use the correct slope for the line you are asked to find, and apply the "parallel" condition correctly.

Question 12. Find a point on the curve \( y = (x - 3)^2 \) where the tangent is parallel to the chord joining (3, 0) and (4, 1).
Answer:
The given equation of the curve is \( y = (x - 3)^2 \).
First, find the slope of the chord joining the points (3, 0) and (4, 1). The formula for the slope between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \frac{y_2 - y_1}{x_2 - x_1} \).
\( m_{\text{chord}} = \frac{1 - 0}{4 - 3} = \frac{1}{1} = 1 \)
Since the tangent to the curve is parallel to this chord, the slope of the tangent must be equal to the slope of the chord. So, \( m_{\text{tangent}} = 1 \).
Now, differentiate the curve's equation with respect to \( x \) to find the slope of the tangent:
\( \frac{dy}{dx} = 2(x - 3) \times 1 = 2(x - 3) \)
Set the slope of the tangent equal to 1:
\( 2(x - 3) = 1 \)
\( x - 3 = \frac{1}{2} \)
\( x = 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2} \)
Now, substitute \( x = \frac{7}{2} \) back into the original curve equation \( y = (x - 3)^2 \) to find the y-coordinate:
\( y = \left(\frac{7}{2} - 3\right)^2 = \left(\frac{7}{2} - \frac{6}{2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
Thus, the required point on the curve is \( \left(\frac{7}{2}, \frac{1}{4}\right) \).
In simple words: First, calculate the slope of the straight line segment (chord) connecting the two given points. Since the tangent line on the curve is parallel to this chord, its slope will be the same. Next, find the derivative of the curve's equation. Set this derivative equal to the chord's slope and solve for the x-value. Finally, use this x-value in the original curve equation to find the corresponding y-value, which gives you the exact point on the curve.

🎯 Exam Tip: Always calculate the slope of the chord first. Remember that "parallel" means equal slopes. Be careful with fractional arithmetic when solving for x and y coordinates.

Question 13. On the curve \( y = x + \frac { 1 }{ x } \), find the points at which the tangents to the curve are parallel to the x-axis.
Answer:
The given equation of the curve is \( y = x + \frac{1}{x} \).
When the tangent to the curve is parallel to the x-axis, its slope is 0.
First, differentiate the curve's equation with respect to \( x \). Rewrite \( \frac{1}{x} \) as \( x^{-1} \):
\( y = x + x^{-1} \)
\( \frac{dy}{dx} = 1 - 1x^{-2} = 1 - \frac{1}{x^2} \)
Set the slope of the tangent to 0:
\( 1 - \frac{1}{x^2} = 0 \)
\( 1 = \frac{1}{x^2} \)
\( x^2 = 1 \)
\( x = \pm 1 \)
Now, find the corresponding y-coordinates for these x-values from the original curve equation \( y = x + \frac{1}{x} \):
When \( x = 1 \):
\( y = 1 + \frac{1}{1} = 1 + 1 = 2 \)
So, one point is (1, 2).
When \( x = -1 \):
\( y = -1 + \frac{1}{-1} = -1 - 1 = -2 \)
So, the other point is (-1, -2).
Thus, the required points on the curve where the tangents are parallel to the x-axis are (1, 2) and (-1, -2).
In simple words: For a tangent to be parallel to the x-axis, its slope must be zero. So, first find the derivative of the curve's equation. Set this derivative equal to zero and solve for the x-values. Then, take these x-values and plug them back into the original curve equation to find their matching y-values. These (x, y) pairs are the points where the tangent is horizontal.

🎯 Exam Tip: Remember that \( \frac{1}{x} \) is \( x^{-1} \). Applying the power rule for differentiation ( \( \frac{d}{dx} x^n = nx^{n-1} \) ) to \( x^{-1} \) is crucial. Don't forget to find both positive and negative solutions for \( x^2=1 \).

Question 14. Find the equation of the tangent to the curve \(y = \cot^2 x - 2 \cot x + 2\) at \(x = \frac{\pi}{4}\).
Answer: The given equation of the curve is \(y = \cot^2 x - 2 \cot x + 2\).
First, we find the y-coordinate at \(x = \frac{\pi}{4}\):
\(y = \cot^2 (\frac{\pi}{4}) - 2 \cot (\frac{\pi}{4}) + 2\)
\(y = (1)^2 - 2(1) + 2\)
\(y = 1 - 2 + 2 = 1\)
So, the point of tangency is \((\frac{\pi}{4}, 1)\).
Next, we differentiate the curve equation with respect to \(x\) to find the slope of the tangent:
\( \frac{dy}{dx} = \frac{d}{dx}(\cot^2 x - 2 \cot x + 2) \)
\( \frac{dy}{dx} = 2 \cot x (-\operatorname{cosec}^2 x) - 2 (-\operatorname{cosec}^2 x) \)
\( \frac{dy}{dx} = -2 \cot x \operatorname{cosec}^2 x + 2 \operatorname{cosec}^2 x \)
Now, we evaluate the slope at \(x = \frac{\pi}{4}\):
\( (\frac{dy}{dx})_{x=\frac{\pi}{4}} = -2 \cot (\frac{\pi}{4}) \operatorname{cosec}^2 (\frac{\pi}{4}) + 2 \operatorname{cosec}^2 (\frac{\pi}{4}) \)
\( = -2(1)(\sqrt{2})^2 + 2(\sqrt{2})^2 \)
\( = -2(1)(2) + 2(2) \)
\( = -4 + 4 = 0 \)
The slope of the tangent at this point is 0, which means the tangent line is horizontal. A horizontal tangent is a flat line.
Finally, we use the point-slope form to find the equation of the tangent line:
\(y - y_1 = m(x - x_1)\)
\(y - 1 = 0(x - \frac{\pi}{4})\)
\(y - 1 = 0\)
\(y = 1\)
In simple words: First, we found the height of the curve at the given x-point. Then, we used derivatives to find how steep the curve is at that exact point. Since the slope was zero, we knew the tangent line would be flat and straight across, passing through the y-value we found.

🎯 Exam Tip: Remember that if the slope of the tangent is 0, the tangent line is a horizontal line of the form \(y = \text{constant}\). Always calculate the y-coordinate first to get the full point of tangency.

Question 15. The equation of tangent at (2, 3) on the curve \(y^2 = ax^3 + b\) is \(y = 4x - 5\). Find the values of a and b.
Answer: The given curve is \(y^2 = ax^3 + b\) (Equation 1).
The equation of the tangent at point (2, 3) is given as \(y = 4x - 5\).
From the tangent equation \(y = 4x - 5\), we can see that its slope is \(m = 4\). This slope must be equal to the derivative of the curve at the point (2, 3).
Now, we differentiate the curve equation (1) implicitly with respect to \(x\):
\( \frac{d}{dx}(y^2) = \frac{d}{dx}(ax^3 + b) \)
\( 2y \frac{dy}{dx} = 3ax^2 \)
\( \frac{dy}{dx} = \frac{3ax^2}{2y} \)
Next, we evaluate the slope of the curve at the given point (2, 3) by substituting \(x=2\) and \(y=3\):
\( (\frac{dy}{dx})_{(2,3)} = \frac{3a(2)^2}{2(3)} \)
\( = \frac{3a(4)}{6} \)
\( = \frac{12a}{6} \)
\( = 2a \)
Since the tangent line's slope is 4, we set the calculated slope equal to 4:
\( 2a = 4 \)
\( \implies a = \frac{4}{2} \)
\( \implies a = 2 \)
The point (2, 3) lies on the curve \(y^2 = ax^3 + b\). So, we can substitute \(x=2\), \(y=3\), and \(a=2\) into the curve equation to find \(b\):
\( 3^2 = (2)(2)^3 + b \)
\( 9 = 2(8) + b \)
\( 9 = 16 + b \)
\( \implies b = 9 - 16 \)
\( \implies b = -7 \)
Thus, the values are \(a = 2\) and \(b = -7\). It's helpful to remember that if a point lies on both a curve and its tangent, it must satisfy both equations.
In simple words: We used the given tangent line to find its slope. Then, we found the general slope formula for the curve using differentiation. We made these two slopes equal at the given point to find the value of 'a'. Finally, we put the point and the value of 'a' into the curve's equation to find 'b'.

🎯 Exam Tip: When a point lies on both the curve and its tangent, it must satisfy both equations. Also, ensure you perform implicit differentiation correctly if the equation is not explicitly in terms of \(y\).