RBSE Solutions Class 12 Maths Chapter 8 Application of Derivatives Exercise 8.3

Question 1. For curve \( y = x^3 - x \) find slope of tangent at \( x = 2 \).
Answer:
Given the curve equation: \( y = x^3 - x \).
To find the slope of the tangent, we first need to find the derivative of \( y \) with respect to \( x \).
Differentiating \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(x^3 - x) \)
\( \frac{dy}{dx} = 3x^2 - 1 \)
Now, we need to find the slope of the tangent at \( x = 2 \). We will substitute \( x = 2 \) into the derivative:
\( \left(\frac{dy}{dx}\right)_{x=2} = 3(2)^2 - 1 \)
\( = 3(4) - 1 \)
\( = 12 - 1 \)
\( = 11 \)
So, the slope of the tangent at \( x = 2 \) is 11. This value represents how steep the curve is at that specific point.
In simple words: First, we find a formula for how the curve changes (its derivative). Then, we put \( x=2 \) into that formula to get the exact steepness, or slope, of the line that just touches the curve at that point.

🎯 Exam Tip: Remember that finding the derivative \( \frac{dy}{dx} \) gives you the general slope formula for any point on the curve. Substituting a specific x-value into this derivative gives the slope of the tangent at that particular point.

Question 3. Find all points where slope of tangent to the curve \( y = \sqrt{(4x-3)} - 1 \) is \( \frac{2}{3} \).
Answer:
Given the curve equation: \( y = \sqrt{4x-3} - 1 \).
We are looking for points where the slope of the tangent is \( \frac{2}{3} \).
First, find the derivative of \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(\sqrt{4x-3} - 1) \)
\( \frac{dy}{dx} = \frac{1}{2\sqrt{4x-3}} \cdot 4 - 0 \)
\( \frac{dy}{dx} = \frac{2}{\sqrt{4x-3}} \)
Now, we set this derivative equal to the given slope \( \frac{2}{3} \):
\( \frac{2}{\sqrt{4x-3}} = \frac{2}{3} \)
We can cross-multiply or simply equate the denominators since the numerators are equal:
\( \sqrt{4x-3} = 3 \)
To solve for \( x \), we square both sides:
\( (\sqrt{4x-3})^2 = 3^2 \)
\( 4x-3 = 9 \)
\( 4x = 9 + 3 \)
\( 4x = 12 \)
\( x = \frac{12}{4} \)
\( x = 3 \)
Now that we have the \( x \)-coordinate, we need to find the corresponding \( y \)-coordinate by substituting \( x = 3 \) back into the original curve equation:
\( y = \sqrt{4(3)-3} - 1 \)
\( y = \sqrt{12-3} - 1 \)
\( y = \sqrt{9} - 1 \)
\( y = 3 - 1 \)
\( y = 2 \)
So, the required point where the slope of the tangent is \( \frac{2}{3} \) is \( (3, 2) \). This point is unique because a square root function typically has a single positive output.
In simple words: First, we found how fast the curve is changing by using differentiation. We then set this change rate equal to the given slope, \( \frac{2}{3} \), and solved to find the \( x \)-value. Finally, we put the \( x \)-value back into the original curve equation to find the matching \( y \)-value.

🎯 Exam Tip: When finding points where the tangent has a specific slope, always remember to find both the \( x \) and \( y \) coordinates. First calculate \( x \) from the derivative, then substitute it back into the original curve equation to find \( y \).

Question 4. Find all equations of lines which are tangent to the curve \( y + \frac{2}{x-3} = 0 \) and slope of those is 2.
Answer:
Given the curve equation: \( y + \frac{2}{x-3} = 0 \)
We can rewrite this as: \( y = -\frac{2}{x-3} \)
We are given that the slope of the tangent is 2.
First, find the derivative of \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}\left(-2(x-3)^{-1}\right) \)
\( \frac{dy}{dx} = -2 \cdot (-1) (x-3)^{-2} \cdot 1 \)
\( \frac{dy}{dx} = \frac{2}{(x-3)^2} \)
Now, set the derivative equal to the given slope, 2:
\( \frac{2}{(x-3)^2} = 2 \)
Divide both sides by 2:
\( \frac{1}{(x-3)^2} = 1 \)
\( (x-3)^2 = 1 \)
Take the square root of both sides:
\( x-3 = \pm 1 \)
This gives two possible values for \( x \):
\( x-3 = 1 \implies x = 1+3 \implies x = 4 \)
\( x-3 = -1 \implies x = -1+3 \implies x = 2 \)
Now we find the corresponding \( y \)-values for each \( x \) by substituting them into the original curve equation \( y = -\frac{2}{x-3} \).
**Case 1: When \( x = 4 \)**
\( y = -\frac{2}{4-3} = -\frac{2}{1} = -2 \)
So, the first point of tangency is \( (4, -2) \).
The equation of the tangent line at \( (4, -2) \) with slope \( m=2 \) is given by \( y - y_1 = m(x - x_1) \):
\( y - (-2) = 2(x - 4) \)
\( y + 2 = 2x - 8 \)
\( 2x - y - 10 = 0 \)
**Case 2: When \( x = 2 \)**
\( y = -\frac{2}{2-3} = -\frac{2}{-1} = 2 \)
So, the second point of tangency is \( (2, 2) \).
The equation of the tangent line at \( (2, 2) \) with slope \( m=2 \) is given by \( y - y_1 = m(x - x_1) \):
\( y - 2 = 2(x - 2) \)
\( y - 2 = 2x - 4 \)
\( 2x - y - 2 = 0 \)
Hence, the required equations of the tangent lines are \( 2x - y - 10 = 0 \) and \( 2x - y - 2 = 0 \). These two lines are parallel and touch the curve at different points.
In simple words: First, we found the formula for the curve's steepness. We set this equal to the given slope (2) and found two \( x \) values where this is true. For each \( x \)-value, we found its matching \( y \)-value on the curve. Then, using the point and the slope, we wrote the equation for each tangent line.

🎯 Exam Tip: Always remember that an equation like \( (x-a)^2 = k \) often gives two solutions for \( x \) (when \( k>0 \)), leading to two distinct points of tangency and thus two tangent lines. Make sure to find both equations.

Question 5. For curve \( \frac{x^2}{4} + \frac{y^2}{25} = 1 \), find those points where tangent is
(i) Parallel to x-axis
(ii) Parallel to y-axis.
Answer:
Given the curve equation: \( \frac{x^2}{4} + \frac{y^2}{25} = 1 \).
First, we need to find the derivative \( \frac{dy}{dx} \). Differentiate the equation with respect to \( x \):
\( \frac{d}{dx}\left(\frac{x^2}{4}\right) + \frac{d}{dx}\left(\frac{y^2}{25}\right) = \frac{d}{dx}(1) \)
\( \frac{2x}{4} + \frac{2y}{25} \frac{dy}{dx} = 0 \)
\( \frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0 \)
Now, solve for \( \frac{dy}{dx} \):
\( \frac{2y}{25} \frac{dy}{dx} = -\frac{x}{2} \)
\( \frac{dy}{dx} = -\frac{x}{2} \cdot \frac{25}{2y} \)
\( \frac{dy}{dx} = -\frac{25x}{4y} \)

(i) **When the tangent is parallel to the x-axis:**
If a tangent line is parallel to the x-axis, its slope must be 0.
So, we set \( \frac{dy}{dx} = 0 \):
\( -\frac{25x}{4y} = 0 \)
This equation implies that the numerator must be zero (assuming \( y \neq 0 \)):
\( -25x = 0 \implies x = 0 \)
Now, substitute \( x = 0 \) back into the original curve equation \( \frac{x^2}{4} + \frac{y^2}{25} = 1 \) to find the \( y \)-coordinates:
\( \frac{0^2}{4} + \frac{y^2}{25} = 1 \)
\( 0 + \frac{y^2}{25} = 1 \)
\( y^2 = 25 \)
\( y = \pm 5 \)
Hence, the tangents are parallel to the x-axis at points \( (0, 5) \) and \( (0, -5) \). At these points, the curve reaches its maximum and minimum vertical extent.

(ii) **When the tangent is parallel to the y-axis:**
If a tangent line is parallel to the y-axis, its slope is undefined. This happens when the denominator of \( \frac{dy}{dx} \) is zero (and the numerator is not zero).
So, we set the denominator \( 4y = 0 \):
\( 4y = 0 \implies y = 0 \)
Now, substitute \( y = 0 \) back into the original curve equation \( \frac{x^2}{4} + \frac{y^2}{25} = 1 \) to find the \( x \)-coordinates:
\( \frac{x^2}{4} + \frac{0^2}{25} = 1 \)
\( \frac{x^2}{4} + 0 = 1 \)
\( x^2 = 4 \)
\( x = \pm 2 \)
Hence, the tangents are parallel to the y-axis at points \( (2, 0) \) and \( (-2, 0) \). These are the leftmost and rightmost points of the ellipse.
In simple words: First, we found the general slope formula for any point on the curve. For tangents parallel to the x-axis, the slope is zero, so we set the top part of our slope formula to zero to find \( x \). For tangents parallel to the y-axis, the slope is undefined, so we set the bottom part of our slope formula to zero to find \( y \). Then, we use these \( x \) or \( y \) values in the original curve equation to find the full points.

🎯 Exam Tip: For tangents parallel to the x-axis, set \( \frac{dy}{dx} = 0 \). For tangents parallel to the y-axis, the slope is undefined, meaning the denominator of \( \frac{dy}{dx} \) is zero. Always substitute back into the original curve equation to find the corresponding coordinate.

Question 6. For curve \( x = a \sin^3 t, y = b \cos^3 t \) find equation of tangent at \( t = \frac{\pi}{2} \).
Answer:
Given the parametric curve equations:
\( x = a \sin^3 t \)
\( y = b \cos^3 t \)
We need to find the equation of the tangent at \( t = \frac{\pi}{2} \).
First, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
\( \frac{dx}{dt} = \frac{d}{dt}(a \sin^3 t) = a \cdot 3 \sin^2 t \cos t \)
\( \frac{dy}{dt} = \frac{d}{dt}(b \cos^3 t) = b \cdot 3 \cos^2 t (-\sin t) = -3b \cos^2 t \sin t \)
Now, find \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \):
\( \frac{dy}{dx} = \frac{-3b \cos^2 t \sin t}{3a \sin^2 t \cos t} \)
\( \frac{dy}{dx} = -\frac{b \cos t}{a \sin t} \)
\( \frac{dy}{dx} = -\frac{b}{a} \cot t \)
Next, find the slope of the tangent at \( t = \frac{\pi}{2} \):
\( \left(\frac{dy}{dx}\right)_{t=\frac{\pi}{2}} = -\frac{b}{a} \cot\left(\frac{\pi}{2}\right) \)
Since \( \cot\left(\frac{\pi}{2}\right) = \frac{\cos(\pi/2)}{\sin(\pi/2)} = \frac{0}{1} = 0 \):
\( \left(\frac{dy}{dx}\right)_{t=\frac{\pi}{2}} = -\frac{b}{a} \cdot 0 = 0 \)
So, the slope of the tangent at \( t = \frac{\pi}{2} \) is 0. This indicates a horizontal tangent.
Now, find the coordinates \( (x, y) \) at \( t = \frac{\pi}{2} \):
\( x = a \sin^3\left(\frac{\pi}{2}\right) = a (1)^3 = a \)
\( y = b \cos^3\left(\frac{\pi}{2}\right) = b (0)^3 = 0 \)
So, the point of tangency is \( (a, 0) \).
The equation of the tangent line at \( (a, 0) \) with slope \( m=0 \) is given by \( y - y_1 = m(x - x_1) \):
\( y - 0 = 0(x - a) \)
\( y = 0 \)
This means the tangent line is the x-axis itself. This is a common occurrence when a curve reaches a peak or valley.
In simple words: First, we found how \( x \) and \( y \) change with \( t \), and used these to find the slope of the tangent line. We then put \( t = \frac{\pi}{2} \) into the slope formula to find the exact steepness, which was 0. Finally, we found the \( x \) and \( y \) coordinates at \( t = \frac{\pi}{2} \) and used the point and slope to write the tangent line equation.

🎯 Exam Tip: For parametric equations, remember to use the chain rule \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Always evaluate both the coordinates and the slope at the given parameter value before writing the tangent equation.

Question 7. For curves \( y = \sin^2 x \), find equation of normal at \( \left(\frac{\pi}{3}, \frac{3}{4}\right) \).
Answer:
Given the curve equation: \( y = \sin^2 x \).
We need to find the equation of the normal at the point \( \left(\frac{\pi}{3}, \frac{3}{4}\right) \).
First, find the derivative \( \frac{dy}{dx} \) to get the slope of the tangent:
\( \frac{dy}{dx} = \frac{d}{dx}(\sin^2 x) \)
Using the chain rule, \( \frac{dy}{dx} = 2 \sin x \cos x \)
We know that \( 2 \sin x \cos x = \sin(2x) \), so:
\( \frac{dy}{dx} = \sin(2x) \)
Now, find the slope of the tangent at the given point \( x = \frac{\pi}{3} \):
\( m_{tangent} = \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{3}} = \sin\left(2 \cdot \frac{\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) \)
Since \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\( m_{tangent} = \frac{\sqrt{3}}{2} \)
The slope of the normal, \( m_{normal} \), is the negative reciprocal of the slope of the tangent:
\( m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} \)
Now, we use the point-slope form \( y - y_1 = m_{normal}(x - x_1) \) to find the equation of the normal at \( \left(\frac{\pi}{3}, \frac{3}{4}\right) \) with slope \( -\frac{2}{\sqrt{3}} \):
\( y - \frac{3}{4} = -\frac{2}{\sqrt{3}}\left(x - \frac{\pi}{3}\right) \)
To remove fractions and simplify, multiply the entire equation by \( 4\sqrt{3} \):
\( 4\sqrt{3}\left(y - \frac{3}{4}\right) = 4\sqrt{3}\left(-\frac{2}{\sqrt{3}}\left(x - \frac{\pi}{3}\right)\right) \)
\( 4\sqrt{3}y - 3\sqrt{3} = -8\left(x - \frac{\pi}{3}\right) \)
\( 4\sqrt{3}y - 3\sqrt{3} = -8x + \frac{8\pi}{3} \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( 8x + 4\sqrt{3}y - 3\sqrt{3} - \frac{8\pi}{3} = 0 \)
Multiply by 3 to clear the fraction:
\( 24x + 12\sqrt{3}y - 9\sqrt{3} - 8\pi = 0 \)
Alternatively, from \( 4\sqrt{3}y - 3\sqrt{3} = -8x + \frac{8\pi}{3} \), we can write:
\( 8x + 4\sqrt{3}y = 3\sqrt{3} + \frac{8\pi}{3} \)
\( 8x + 4\sqrt{3}y = \frac{9\sqrt{3} + 8\pi}{3} \)
\( 24x + 12\sqrt{3}y = 9\sqrt{3} + 8\pi \)
The equation \( 2x + \sqrt{3}y = \frac{2\pi}{3} + \frac{3\sqrt{3}}{4} \) shown in the source implies a different simplification path. Let's re-trace the source's simplification from \( y - \frac{3}{4} = -\frac{2}{\sqrt{3}}\left(x - \frac{\pi}{3}\right) \).
Multiply through by \( \sqrt{3} \): \( \sqrt{3}y - \frac{3\sqrt{3}}{4} = -2x + \frac{2\pi}{3} \)
Move \( x \) to the left: \( 2x + \sqrt{3}y = \frac{2\pi}{3} + \frac{3\sqrt{3}}{4} \)
This matches the source's final form, and is a perfectly valid representation of the line. The normal line is perpendicular to the tangent at the point of contact.
In simple words: First, we found the slope of the tangent line by differentiating the curve. Then, we found the slope of the normal line by taking the negative inverse of the tangent's slope. Finally, using this normal slope and the given point, we wrote the equation for the normal line.

🎯 Exam Tip: Always remember that the slope of the normal is \( -\frac{1}{m_{tangent}} \). Be careful with trigonometric values and simplifying the final equation of the line, especially when dealing with fractions involving \( \pi \) and square roots.

Question 8. Find equation of tangent and normal following curves, at given points.
(d) \( y^2 = 4ax \), at \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \)
Answer:
Given the curve: \( y^2 = 4ax \).
Point of tangency: \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \).
First, find the derivative \( \frac{dy}{dx} \). Differentiate \( y^2 = 4ax \) with respect to \( x \):
\( 2y \frac{dy}{dx} = 4a \)
\( \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \)
Now, find the slope of the tangent at the given point \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \). Substitute \( y = \frac{2a}{m} \):
\( m_{tangent} = \left(\frac{dy}{dx}\right)_{\left(\frac{a}{m^2}, \frac{2a}{m}\right)} = \frac{2a}{\frac{2a}{m}} = m \)
So, the slope of the tangent is \( m \).
**Equation of Tangent:**
Using the point-slope form \( y - y_1 = m_{tangent}(x - x_1) \) with point \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \) and slope \( m \):
\( y - \frac{2a}{m} = m\left(x - \frac{a}{m^2}\right) \)
\( y - \frac{2a}{m} = mx - \frac{a}{m} \)
Multiply the entire equation by \( m \) to clear the denominators:
\( my - 2a = m^2 x - a \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( m^2 x - my + a = 0 \)
This is the equation of the tangent.
**Equation of Normal:**
The slope of the normal, \( m_{normal} \), is the negative reciprocal of the tangent's slope:
\( m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{m} \)
Using the point-slope form \( y - y_1 = m_{normal}(x - x_1) \) with point \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \) and slope \( -\frac{1}{m} \):
\( y - \frac{2a}{m} = -\frac{1}{m}\left(x - \frac{a}{m^2}\right) \)
Multiply the entire equation by \( m^2 \) to clear the denominators:
\( m^2 y - 2am = -m\left(x - \frac{a}{m^2}\right) \)
\( m^2 y - 2am = -mx + \frac{a}{m} \cdot m = -mx + a \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( mx + m^2 y - 2am - a = 0 \)
\( mx + m^2 y - a(2m + 1) = 0 \)
So, the equations are \( m^2 x - my + a = 0 \) for the tangent and \( mx + m^2 y - a(2m+1) = 0 \) for the normal. These equations describe lines touching and being perpendicular to the parabola at the specified point.
In simple words: First, we found the slope of the curve at the given point. This is the tangent's slope. Then, using this slope and the given point, we wrote the tangent's equation. For the normal, we found its slope by taking the negative flip of the tangent's slope, and then used that slope with the same point to write the normal's equation.

🎯 Exam Tip: When dealing with parametric forms or specific points, ensure you correctly substitute both \( x \) and \( y \) values (or the parameter \( m \)) into the derivative to find the slope at that exact point. Be meticulous with algebraic simplification.

Question 8. Find equation of tangent and normal following curves, at given points.
(e) \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), at \( (a \sec \theta, b \tan \theta) \)
Answer:
Given the curve (a hyperbola): \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Point of tangency: \( (a \sec \theta, b \tan \theta) \).
First, find the derivative \( \frac{dy}{dx} \). Differentiate the equation with respect to \( x \):
\( \frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \)
\( -\frac{2y}{b^2} \frac{dy}{dx} = -\frac{2x}{a^2} \)
\( \frac{dy}{dx} = \frac{2x}{a^2} \cdot \frac{b^2}{2y} = \frac{b^2 x}{a^2 y} \)
Now, find the slope of the tangent at the given point \( (a \sec \theta, b \tan \theta) \). Substitute \( x = a \sec \theta \) and \( y = b \tan \theta \):
\( m_{tangent} = \left(\frac{dy}{dx}\right)_{(a \sec \theta, b \tan \theta)} = \frac{b^2 (a \sec \theta)}{a^2 (b \tan \theta)} \)
\( m_{tangent} = \frac{b^2 a \sec \theta}{a^2 b \tan \theta} = \frac{b \sec \theta}{a \tan \theta} \)
**Equation of Tangent:**
Using the point-slope form \( y - y_1 = m_{tangent}(x - x_1) \) with point \( (a \sec \theta, b \tan \theta) \) and slope \( \frac{b \sec \theta}{a \tan \theta} \):
\( y - b \tan \theta = \frac{b \sec \theta}{a \tan \theta} (x - a \sec \theta) \)
Multiply both sides by \( a \tan \theta \) to clear the denominator:
\( a \tan \theta (y - b \tan \theta) = b \sec \theta (x - a \sec \theta) \)
\( ay \tan \theta - ab \tan^2 \theta = bx \sec \theta - ab \sec^2 \theta \)
Rearrange terms to get the standard form of the tangent to a hyperbola:
\( bx \sec \theta - ay \tan \theta = ab \sec^2 \theta - ab \tan^2 \theta \)
\( bx \sec \theta - ay \tan \theta = ab (\sec^2 \theta - \tan^2 \theta) \)
Recall the identity \( \sec^2 \theta - \tan^2 \theta = 1 \):
\( bx \sec \theta - ay \tan \theta = ab (1) \)
\( bx \sec \theta - ay \tan \theta = ab \)
Divide by \( ab \):
\( \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1 \)
This is the equation of the tangent.
**Equation of Normal:**
The slope of the normal, \( m_{normal} \), is the negative reciprocal of the tangent's slope:
\( m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{\frac{b \sec \theta}{a \tan \theta}} = -\frac{a \tan \theta}{b \sec \theta} \)
Using the point-slope form \( y - y_1 = m_{normal}(x - x_1) \) with point \( (a \sec \theta, b \tan \theta) \) and slope \( -\frac{a \tan \theta}{b \sec \theta} \):
\( y - b \tan \theta = -\frac{a \tan \theta}{b \sec \theta} (x - a \sec \theta) \)
Multiply both sides by \( b \sec \theta \) to clear the denominator:
\( b \sec \theta (y - b \tan \theta) = -a \tan \theta (x - a \sec \theta) \)
\( by \sec \theta - b^2 \sec \theta \tan \theta = -ax \tan \theta + a^2 \sec \theta \tan \theta \)
Rearrange terms:
\( ax \tan \theta + by \sec \theta = a^2 \sec \theta \tan \theta + b^2 \sec \theta \tan \theta \)
\( ax \tan \theta + by \sec \theta = \tan \theta \sec \theta (a^2 + b^2) \)
Divide by \( \sec \theta \) (or \( \tan \theta \sec \theta \)) if \( \sec \theta \neq 0 \) and \( \tan \theta \neq 0 \).
A common form is \( \frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2 + b^2 \), which implies multiplying by \( \cos \theta \) and \( \cot \theta \):
\( ax \cos \theta + by \cot \theta = a^2 + b^2 \)
This is the equation of the normal. These equations precisely define the lines that touch and are perpendicular to the hyperbola at a given point defined by the parameter \( \theta \).
In simple words: First, we calculated the slope of the line that touches the hyperbola at the given point. This is the tangent's slope. Using this slope and the given point, we wrote the tangent's equation. For the normal, we found its slope by taking the negative inverse of the tangent's slope, and then used that new slope with the same point to write the normal's equation.

🎯 Exam Tip: For equations of tangent and normal to conic sections (like hyperbolas), remember the standard forms. The identity \( \sec^2 \theta - \tan^2 \theta = 1 \) is crucial for simplifying the tangent equation to a concise form. Double-check your algebraic manipulations for each step.

Question 8. Find equation of tangent and normal following curves, at given points.
(f) \( y = 2x^2 - 3x - 1 \), at \( (1, -2) \)
Answer:
Given the curve: \( y = 2x^2 - 3x - 1 \).
Point of tangency: \( (1, -2) \).
First, find the derivative \( \frac{dy}{dx} \) to get the slope of the tangent:
\( \frac{dy}{dx} = \frac{d}{dx}(2x^2 - 3x - 1) \)
\( \frac{dy}{dx} = 4x - 3 \)
Now, find the slope of the tangent at the given point \( (1, -2) \). Substitute \( x = 1 \):
\( m_{tangent} = \left(\frac{dy}{dx}\right)_{x=1} = 4(1) - 3 = 4 - 3 = 1 \)
So, the slope of the tangent is 1.
**Equation of Tangent:**
Using the point-slope form \( y - y_1 = m_{tangent}(x - x_1) \) with point \( (1, -2) \) and slope \( m=1 \):
\( y - (-2) = 1(x - 1) \)
\( y + 2 = x - 1 \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( x - y - 3 = 0 \)
This is the equation of the tangent.
**Equation of Normal:**
The slope of the normal, \( m_{normal} \), is the negative reciprocal of the tangent's slope:
\( m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{1} = -1 \)
Using the point-slope form \( y - y_1 = m_{normal}(x - x_1) \) with point \( (1, -2) \) and slope \( m=-1 \):
\( y - (-2) = -1(x - 1) \)
\( y + 2 = -x + 1 \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( x + y + 1 = 0 \)
So, the equations are \( x - y - 3 = 0 \) for the tangent and \( x + y + 1 = 0 \) for the normal. These lines show the local direction and perpendicular direction of the curve at \( (1, -2) \).
In simple words: We first found the slope of the tangent at the given point by differentiating the curve. Then, we used this slope and the point to write the tangent's equation. For the normal, we used the negative opposite of the tangent's slope and the same point to write its equation.

🎯 Exam Tip: When the slope of the tangent is 1, the slope of the normal is -1. This is a common pair. Always double check your arithmetic, especially with negative signs, when calculating slopes and forming equations.

Question 8. Find equation of tangent and normal following curves, at given points.
(g) \( x = at^2, y = 2at \), \( t = 1 \)
Answer:
Given the parametric curve equations:
\( x = at^2 \)
\( y = 2at \)
We need to find the equation of the tangent and normal at \( t = 1 \).
First, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
\( \frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at \)
\( \frac{dy}{dt} = \frac{d}{dt}(2at) = 2a \)
Now, find \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \):
\( \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t} \)
Next, find the slope of the tangent at \( t = 1 \):
\( m_{tangent} = \left(\frac{dy}{dx}\right)_{t=1} = \frac{1}{1} = 1 \)
So, the slope of the tangent is 1.
Now, find the coordinates \( (x, y) \) at \( t = 1 \):
\( x = a(1)^2 = a \)
\( y = 2a(1) = 2a \)
So, the point of tangency is \( (a, 2a) \).
**Equation of Tangent:**
Using the point-slope form \( y - y_1 = m_{tangent}(x - x_1) \) with point \( (a, 2a) \) and slope \( m=1 \):
\( y - 2a = 1(x - a) \)
\( y - 2a = x - a \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( x - y + a = 0 \)
This is the equation of the tangent.
**Equation of Normal:**
The slope of the normal, \( m_{normal} \), is the negative reciprocal of the tangent's slope:
\( m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{1} = -1 \)
Using the point-slope form \( y - y_1 = m_{normal}(x - x_1) \) with point \( (a, 2a) \) and slope \( m=-1 \):
\( y - 2a = -1(x - a) \)
\( y - 2a = -x + a \)
Rearrange into the standard form \( Ax + By + C = 0 \):
\( x + y - 3a = 0 \)
So, the equations are \( x - y + a = 0 \) for the tangent and \( x + y - 3a = 0 \) for the normal. This is a common result for the parabola \( y^2=4ax \) in parametric form, where \( t \) is the parameter.
In simple words: We first found how \( x \) and \( y \) change with \( t \), then used these to find the slope of the tangent. At \( t=1 \), we found the slope and the point on the curve. With these, we wrote the equation for the tangent line. For the normal, we used the negative inverse of the tangent's slope and the same point to write its equation.

🎯 Exam Tip: For parametric equations, remember that \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). After finding \( \frac{dy}{dx} \) in terms of \( t \), substitute the given value of \( t \) to get the numerical slope. Always calculate the point \( (x, y) \) using the given \( t \) value before writing the equation of the line.

Question 8. Find equation of tangent and normal following curves, at given points.
(h) \( x = \theta + \sin \theta, y = 1 - \cos \theta \), at \( \theta = \frac{\pi}{2} \)
Answer:
Given the parametric curve equations:
\( x = \theta + \sin \theta \)
\( y = 1 - \cos \theta \)
We need to find the equation of the tangent and normal at \( \theta = \frac{\pi}{2} \).
First, find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \):
\( \frac{dx}{d\theta} = \frac{d}{d\theta}(\theta + \sin \theta) = 1 + \cos \theta \)
\( \frac{dy}{d\theta} = \frac{d}{d\theta}(1 - \cos \theta) = 0 - (-\sin \theta) = \sin \theta \)
Now, find \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \):
\( \frac{dy}{dx} = \frac{\sin \theta}{1 + \cos \theta} \)
Next, find the slope of the tangent at \( \theta = \frac{\pi}{2} \):
\( m_{tangent} = \left(\frac{dy}{dx}\right)_{\theta=\frac{\pi}{2}} = \frac{\sin(\pi/2)}{1 + \cos(\pi/2)} \)
Since \( \sin(\pi/2) = 1 \) and \( \cos(\pi/2) = 0 \):
\( m_{tangent} = \frac{1}{1 + 0} = 1 \)
So, the slope of the tangent is 1.
Now, find the coordinates \( (x, y) \) at \( \theta = \frac{\pi}{2} \):
\( x = \frac{\pi}{2} + \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + 1 \)
\( y = 1 - \cos\left(\frac{\pi}{2}\right) = 1 - 0 = 1 \)
So, the point of tangency is \( \left(\frac{\pi}{2} + 1, 1\right) \).
**Equation of Tangent:**
Using the point-slope form \( y - y_1 = m_{tangent}(x - x_1) \) with point \( \left(\frac{\pi}{2} + 1, 1\right) \) and slope \( m=1 \):
\( y - 1 = 1\left(x - \left(\frac{\pi}{2} + 1\right)\right) \)
\( y - 1 = x - \frac{\pi}{2} - 1 \)
Rearrange into a simplified form:
\( x - y - \frac{\pi}{2} = 0 \)
\( x - y = \frac{\pi}{2} \)
This is the equation of the tangent.
**Equation of Normal:**
The slope of the normal, \( m_{normal} \), is the negative reciprocal of the tangent's slope:
\( m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{1} = -1 \)
Using the point-slope form \( y - y_1 = m_{normal}(x - x_1) \) with point \( \left(\frac{\pi}{2} + 1, 1\right) \) and slope \( m=-1 \):
\( y - 1 = -1\left(x - \left(\frac{\pi}{2} + 1\right)\right) \)
\( y - 1 = -x + \frac{\pi}{2} + 1 \)
Rearrange into a simplified form:
\( x + y = \frac{\pi}{2} + 2 \)
So, the equations are \( x - y = \frac{\pi}{2} \) for the tangent and \( x + y = \frac{\pi}{2} + 2 \) for the normal. These lines show the orientation of the curve at this specific point.
In simple words: First, we found how \( x \) and \( y \) change with \( \theta \), and then calculated the slope of the tangent at \( \theta = \frac{\pi}{2} \). We also found the exact point on the curve at this \( \theta \). Using this slope and point, we wrote the tangent line's equation. For the normal, we used the negative inverse of the tangent's slope and the same point to write its equation.

🎯 Exam Tip: Pay close attention to trigonometric values for standard angles like \( \frac{\pi}{2} \). A slope of 1 or -1 often simplifies the equation significantly. Always write down the point of tangency clearly before applying the point-slope formula.

Question 8. (h) For curve \( x = \theta + \sin \theta \), \( y = 1 - \cos \theta \) find the equation of tangent and normal at \( \theta = \frac{\pi}{2} \).
Answer: We need to find the point on the curve and the slope of the tangent at \( \theta = \frac{\pi}{2} \). First, find the coordinates \( (x, y) \) by substituting \( \theta = \frac{\pi}{2} \) into the given equations: \( x = \theta + \sin \theta \) \( x = \frac{\pi}{2} + \sin \frac{\pi}{2} = \frac{\pi}{2} + 1 \)
\( y = 1 - \cos \theta \) \( y = 1 - \cos \frac{\pi}{2} = 1 - 0 = 1 \) So, the point of tangency on the curve is \( \left( \frac{\pi}{2} + 1, 1 \right) \). This is the specific point where the tangent and normal lines touch the curve. Next, we calculate the derivatives to find the slope \( \frac{dy}{dx} \). \( \frac{dx}{d\theta} = \frac{d}{d\theta} (\theta + \sin \theta) = 1 + \cos \theta \)
\( \frac{dy}{d\theta} = \frac{d}{d\theta} (1 - \cos \theta) = \sin \theta \)
\( \implies \) The slope of the tangent is \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\sin \theta}{1 + \cos \theta} \) Now, substitute \( \theta = \frac{\pi}{2} \) to get the slope at the specific point: \( m_T = \frac{dy}{dx}|_{\theta=\frac{\pi}{2}} = \frac{\sin \frac{\pi}{2}}{1 + \cos \frac{\pi}{2}} = \frac{1}{1 + 0} = 1 \) Equation of Tangent: Use the point-slope formula \( y - y_1 = m_T(x - x_1) \). Using \( (x_1, y_1) = \left( \frac{\pi}{2} + 1, 1 \right) \) and \( m_T = 1 \): \( y - 1 = 1 \left( x - \left( \frac{\pi}{2} + 1 \right) \right) \) \( y - 1 = x - \frac{\pi}{2} - 1 \) \( \implies x - y - \frac{\pi}{2} = 0 \) This is the equation for the line that just touches the curve at one specific point. Equation of Normal: The slope of the normal (\( m_N \)) is the negative reciprocal of the tangent's slope. \( m_N = -\frac{1}{m_T} = -\frac{1}{1} = -1 \) Using the point-slope formula \( y - y_1 = m_N(x - x_1) \): \( y - 1 = -1 \left( x - \left( \frac{\pi}{2} + 1 \right) \right) \) \( y - 1 = -x + \frac{\pi}{2} + 1 \) \( \implies x + y = 2 + \frac{\pi}{2} \) This line is perpendicular to the tangent at the same point on the curve.
In simple words: First, we find the exact spot on the curve where we want the lines. Then, we find how steep the curve is at that spot using calculus. This steepness gives us the slope of the tangent line. The normal line is simply the one that crosses the tangent line at a perfect right angle.

🎯 Exam Tip: Remember to calculate both the coordinates of the point and the slope correctly before applying the tangent and normal line formulas, as small errors can propagate.