CBSE Class 12 Mathematics HOTs Application of Derivatives Set 04

Question. For the curve \( y = 3x^2 + 4x \), find the slope of the tangent to the curve at the point whose \( x \)-coordinate is -2.
Answer: The given curve is \( y = 3x^2 + 4x \).
\( \Rightarrow \frac{dy}{dx} = 6x + 4 \).
\( \therefore \) Slope of tangent when \( x \)-coordinate is -2, is:
\( \left(\frac{dy}{dx}\right)_{x=-2} = 6(-2) + 4 = -8 \).

Question. Show that the equation of normal at any point \( t \) on the curve \( x = 3\cos t - \cos^3 t \) and \( y = 3\sin t - \sin^3 t \) is \( 4(y\cos^3 t - x\sin^3 t) = 3\sin 4t \).
Answer: \( x = 3\cos t - \cos^3 t \) and \( y = 3\sin t - \sin^3 t \)
Now, \( \frac{dx}{dt} = -3\sin t + 3\cos^2 t\sin t = -3\sin t(1 - \cos^2 t) = -3\sin^3 t \).
Also, \( \frac{dy}{dt} = 3\cos t - 3\sin^2 t\cos t = 3\cos t(1 - \sin^2 t) = 3\cos^3 t \).
So, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos^3 t}{-3\sin^3 t} = -\frac{\cos^3 t}{\sin^3 t} \).
Slope of normal \( = \frac{-1}{dy/dx} = \frac{\sin^3 t}{\cos^3 t} \).
Required equation of normal is:
\( y - (3\sin t - \sin^3 t) = \frac{\sin^3 t}{\cos^3 t} [x - (3\cos t - \cos^3 t)] \)
\( \Rightarrow y\cos^3 t - 3\sin t\cos^3 t + \sin^3 t\cos^3 t = x\sin^3 t - 3\cos t\sin^3 t + \sin^3 t\cos^3 t \)
\( \Rightarrow y\cos^3 t - x\sin^3 t = 3\sin t\cos^t(\cos^2 t - \sin^2 t) = \frac{3\sin 2t \cdot \cos 2t}{2} \)
\( \Rightarrow y\cos^3 t - x\sin^3 t = \frac{3}{2}\frac{\sin 4t}{2} \)
\( \Rightarrow 4(y\cos^3 t - x\sin^3 t) = 3\sin 4t \). Hence proved.

Question. 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: We have, \( y^2 = ax^3 + b \)
Differentiating w.r.t. \( x \), we get
\( 2y\frac{dy}{dx} = 3ax^2 \Rightarrow \frac{dy}{dx} = \frac{3ax^2}{2y} \).
\( \Rightarrow \left(\frac{dy}{dx}\right)_{(2,3)} = \frac{3a(2)^2}{2(3)} = 2a \).
So, equation of tangent at the point \( (2, 3) \) is:
\( y - 3 = 2a(x - 2) \Rightarrow y = 2ax - 4a + 3 \) ...(i)
But we are given that equation of tangent at \( (2, 3) \) is:
\( y = 4x - 5 \) ...(ii)
On comparing (i) and (ii), we get
\( 2a = 4 \Rightarrow a = 2 \).
Since point \( (2, 3) \) lies on the curve \( y^2 = ax^3 + b \),
\( \therefore (3)^2 = (2)^3 a + b \Rightarrow 9 = 8a + b \)
\( \Rightarrow 9 = 8 \times 2 + b \Rightarrow b = -7 \).

Question. Find the angle of intersection of the curves \( y^2 = 4ax \) and \( x^2 = 4by \).
Answer: The given curves are \( y^2 = 4ax \) ...(i) and \( x^2 = 4by \) ...(ii).
Solving (i) and (ii), we get:
\( \left(\frac{x^2}{4b}\right)^2 = 4ax \Rightarrow x^4 - 64ab^2x = 0 \Rightarrow x(x^3 - 64ab^2) = 0 \Rightarrow x = 0, 4a^{1/3}b^{2/3} \).
When \( x = 0, y = 0 \).
When \( x = 4a^{1/3}b^{2/3}, y = \frac{(4a^{1/3}b^{2/3})^2}{4b} = 4a^{2/3}b^{1/3} \).
Thus, the given curves intersect at \( (0, 0) \) and \( \left(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3}\right) \).
At \( (0, 0) \) the angle between the curves is \( 90^\circ \).
Differentiating (i) with respect to \( x \), we get:
\( 2y\frac{dy}{dx} = 4a \Rightarrow \frac{dy}{dx} = \frac{2a}{y} \).
\( \Rightarrow m_1 = \left(\frac{dy}{dx}\right)_{(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})} = \frac{2a}{4a^{2/3}b^{1/3}} = \frac{1}{2}\left(\frac{a}{b}\right)^{1/3} \).
Differentiating (ii) with respect to \( x \), we get:
\( 2x = 4b\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{2b} \).
\( \Rightarrow m_2 = \left(\frac{dy}{dx}\right)_{(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})} = \frac{4a^{1/3}b^{2/3}}{2b} = 2\left(\frac{a}{b}\right)^{1/3} \).
Let \( \theta \) be the angle between the two curves. Then,
\( \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \left|\frac{\frac{1}{2}\left(\frac{a}{b}\right)^{1/3} - 2\left(\frac{a}{b}\right)^{1/3}}{1 + \frac{1}{2}\left(\frac{a}{b}\right)^{1/3} \times 2\left(\frac{a}{b}\right)^{1/3}}\right| = \left|\frac{-\frac{3}{2}\left(\frac{a}{b}\right)^{1/3}}{1 + \left(\frac{a}{b}\right)^{2/3}}\right| = \left|\frac{-3a^{1/3}b^{1/3}}{2(a^{2/3} + b^{2/3})}\right| \).
\( \Rightarrow \theta = \tan^{-1}\left|\frac{-3(ab)^{1/3}}{2(a^{2/3} + b^{2/3})}\right| \).

Question. Find the point on the curve \( 9y^2 = x^3 \), where the normal to the curve makes equal intercepts on the axes.
Answer: We have, \( 9y^2 = x^3 \) ...(i)
Differentiating (i) w.r.t. \( x \), we get:
\( 18y\frac{dy}{dx} = 3x^2 \Rightarrow \frac{dy}{dx} = \frac{x^2}{6y} \).
Let \( P(x_1, y_1) \) be the point on (i) where the normal makes equal intercepts on the axes.
\( \therefore 9y_1^2 = x_1^3 \) ...(ii)
Now, slope of tangent at \( (x_1, y_1) \) is \( \frac{x_1^2}{6y_1} \).
\( \therefore \) Slope of normal at \( (x_1, y_1) \) is \( -\frac{6y_1}{x_1^2} \).
Since the normal makes equal intercepts on the axes,
\( \therefore \) Its slope \( = \pm 1 \Rightarrow \frac{6y_1}{x_1^2} = \pm 1 \) ...(iii)
Now from (ii) and (iii), we get \( 9\left(\pm \frac{x_1^2}{6}\right)^2 = x_1^3 \Rightarrow \frac{x_1^4}{4} = x_1^3 \Rightarrow x_1 = 4 \).
Putting \( x_1 = 4 \) in (iii), we get \( y_1 = \pm \frac{16}{6} = \pm \frac{8}{3} \).
Hence, the points are \( \left(4, \pm\frac{8}{3}\right) \).

Question. Find the equation of the tangent and normal to the curve \( x = a\sin^3\theta \) and \( y = a\cos^3\theta \) at \( \theta = \frac{\pi}{4} \).
Answer: We have, \( x = a\sin^3\theta; y = a\cos^3\theta \).
\( \frac{dx}{d\theta} = 3a\sin^2\theta\cos\theta \) and \( \frac{dy}{d\theta} = 3a\cos^2\theta(-\sin\theta) \).
\( \therefore \frac{dy}{dx} = \frac{-3a\cos^2\theta\sin\theta}{3a\sin^2\theta\cos\theta} = -\frac{\cos\theta}{\sin\theta} = -\cot\theta \).
\( \therefore \left[\frac{dy}{dx}\right]_{\theta = \frac{\pi}{4}} = -\cot\frac{\pi}{4} = -1 \).
Hence, the equation of the tangent at \( \theta = \frac{\pi}{4} \) is:
\( y - a\cos^3\theta = (-1)(x - a\sin^3\theta) \)
\( \Rightarrow x + y = a\left\{\left(\frac{1}{\sqrt{2}}\right)^3 + \left(\frac{1}{\sqrt{2}}\right)^3\right\} = \frac{a}{\sqrt{2}} \)
\( \Rightarrow \sqrt{2}x + \sqrt{2}y = a \),
and the equation of the normal is:
\( y - a\cos^3\theta = \frac{-1}{(-1)}(x - a\sin^3\theta) \)
\( \therefore y - a\cos^3\theta = (x - a\sin^3\theta) \)
\( \Rightarrow x - y = a\left\{\left(\frac{1}{\sqrt{2}}\right)^3 - \left(\frac{1}{\sqrt{2}}\right)^3\right\} \)
\( \Rightarrow x - y = 0 \).

Question. Find the equations of the tangent and normal to the curve \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at the point \( (\sqrt{2}a, b) \).
Answer: The given curve is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Differentiating w.r.t. \( x \), we get:
\[ \frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x/a^2}{y/b^2} = \frac{b^2 x}{a^2 y} \]
At \( P(\sqrt{2}a, b) \),
Slope of the tangent \( = \frac{b^2 \sqrt{2}a}{a^2 b} = \sqrt{2}\frac{b}{a} \)
and slope of the normal \( = -\frac{a}{b \sqrt{2}} \).
Therefore, Equation of the tangent at \( P \) is
\( y - b = \sqrt{2}\frac{b}{a}(x - \sqrt{2}a) \implies y = \sqrt{2}\frac{b}{a}x - b \)
Also equation of the normal at \( P \) is
\( y - b = -\frac{a}{b\sqrt{2}}(x - \sqrt{2}a) \implies y = -\frac{ax}{\sqrt{2}b} + b + \frac{a^2}{b} \)

Question. Find the point on the curve \( y = x^3 - 11x + 5 \) at which the equation of tangent is \( y = x - 11 \).
Answer: \( y = x^3 - 11x + 5 \) ...(i)
Differentiating (i) w.r.t. \( x \), we get
\( \frac{dy}{dx} = 3x^2 - 11 = \text{slope of tangent} \)
Also, equation of tangent is \( y = x - 11 \)
Therefore, its slope = 1.
So \( 3x^2 - 11 = 1 \implies x^2 = 4 \implies x = \pm 2 \).
Putting the values of \( x \) in (i), we get
\( y = 2^3 - 11(2) + 5 = 8 - 22 + 5 = -9 \)
\( y = (-2)^3 - 11(-2) + 5 = -8 + 22 + 5 = 19 \)
So points are \( (2, -9) \) and \( (-2, 19) \).
But only \( (2, -9) \) satisfies the equation of tangent.
So required point is \( (2, -9) \).

Question. Show that the equation of tangent to the parabola \( y^2 = 4ax \) at \( (x_1, y_1) \) is \( y y_1 = 2a(x + x_1) \).
Answer: The Given parabola is
\( y^2 = 4ax \) ...(i)
\( \implies 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} \)
Therefore, Slope of the tangent to (1) at \( (x_1, y_1) \) is \( \frac{2a}{y_1} \)
Equation of the tangent to (1) at \( (x_1, y_1) \) is
\( y - y_1 = \frac{2a}{y_1}(x - x_1) \)
\( \implies y y_1 - y_1^2 = 2ax - 2ax_1 \)
\( \implies y y_1 = 2ax + (y_1^2 - 2ax_1) \) ...(ii)
Now \( (x_1, y_1) \) lies on (i),
Therefore, \( y_1^2 = 4ax_1 \implies y_1^2 - 2ax_1 = 2ax_1 \) ...(iii)
From (ii) and (iii), we get
\( y y_1 = 2ax + 2ax_1 = 2a(x + x_1) \)
This is the equation of the tangent to (i) at \( (x_1, y_1) \).

Question. Find the points on the curve \( x^2 + y^2 - 2x - 3 = 0 \) at which the tangents are parallel to x-axis.
Answer: The given curve is
\( x^2 + y^2 - 2x - 3 = 0 \) ...(i)
Differentiating with respect to \( x \), we get
\( 2x + 2y \frac{dy}{dx} - 2 = 0 \implies \frac{dy}{dx} = \frac{1-x}{y} \)
Since the tangent is parallel to x-axis,
Therefore, \( \frac{dy}{dx} = 0 \implies \frac{1-x}{y} = 0 \implies x = 1 \)
Putting the value of \( x = 1 \) in (i), we get
\( (1)^2 + y^2 - 2(1) - 3 = 0 \implies y^2 = 4 \implies y = \pm 2 \)
Therefore, the points are \( (1, 2) \) and \( (1, -2) \).

Question. Find the equation of the tangent to the curve \( y = x^4 - 6x^3 + 13x^2 - 10x + 5 \) at the point \( x = 1 \).
Answer: Here, \( y = x^4 - 6x^3 + 13x^2 - 10x + 5 \) ...(i)
When \( x = 1 \), \( y = 1 - 6 + 13 - 10 + 5 = 3 \).
We want to find tangent to (i) at \( P(1, 3) \).
Differentiating (i) w.r.t. \( x \), we get
\( \frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10 \)
Therefore, \( \left(\frac{dy}{dx}\right)_P = 4 - 18 + 26 - 10 = 2 \)
Hence the equation of the tangent to (i) at \( P(1, 3) \) is
\( y - 3 = 2(x - 1) \implies y = 2x + 1 \)

Question. Find the equation of the tangent to the curve \( 4x^2 + 9y^2 = 36 \) at the point \( (3 \cos \theta, 2 \sin \theta) \).
Answer: The given curve is
\( 4x^2 + 9y^2 = 36 \) ...(i)
\( \implies 4 \cdot 2x + 9 \cdot 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{4x}{9y} \)
Let \( P(3 \cos \theta, 2 \sin \theta) \)
Therefore, \( \left[\frac{dy}{dx}\right]_P = -\frac{4 \cdot 3 \cos \theta}{9 \cdot 2 \sin \theta} = -\frac{2 \cos \theta}{3 \sin \theta} \)
Therefore, Equation of the tangents to (i) at \( P \) is
\( y - 2 \sin \theta = -\frac{2 \cos \theta}{3 \sin \theta} (x - 3 \cos \theta) \)
\( \implies 3y \sin \theta - 6 \sin^2 \theta = -2x \cos \theta + 6 \cos^2 \theta \)
\( \implies 2x \cos \theta + 3y \sin \theta = 6 \)

Question. Find the equations of tangents to the curve \( y = (x^2 - 1)(x - 2) \) at the points where the curve cuts the x-axis.
Answer: The given curve is
\( y = (x^2 - 1)(x - 2) \) ...(i)
It meets x-axis, where \( y = 0 \)
Therefore, \( (x^2 - 1)(x - 2) = 0 \implies x = 1, -1, 2 \)
Therefore, We want to find tangents at \( (1, 0), (-1, 0) \) and \( (2, 0) \).
Now from equation (i), we get
\( y = x^3 - 2x^2 - x + 2 \)
\( \implies \frac{dy}{dx} = 3x^2 - 4x - 1 \)
Therefore, \( \left(\frac{dy}{dx}\right)_{x=1} = -2 \); \( \left(\frac{dy}{dx}\right)_{x=-1} = 6 \); \( \left(\frac{dy}{dx}\right)_{x=2} = 3 \)
Now, tangent at \( (1, 0) \): \( y - 0 = -2(x - 1) \implies 2x + y = 2 \)
tangent at \( (-1, 0) \): \( y - 0 = 6(x + 1) \implies y = 6x + 6 \)
tangent at \( (2, 0) \): \( y - 0 = 3(x - 2) \implies y = 3x - 6 \)

Question. Find the equation of tangent to the curve \( x = \sin 3t, y = \cos 2t \) at \( t = \frac{\pi}{4} \).
Answer: The curve is \( x = \sin 3t; y = \cos 2t \)
\( \implies \frac{dx}{dt} = 3 \cos 3t; \frac{dy}{dt} = -2 \sin 2t \)
\( \implies \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\sin 2t}{3\cos 3t} \)
At \( t = \frac{\pi}{4} \),
\( x = \sin \frac{3\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
\( y = \cos 2t = \cos \frac{\pi}{2} = 0 \)
and \( \frac{dy}{dx} = \frac{2 \sin \frac{\pi}{2}}{\cos \frac{3\pi}{4}} = \frac{2 \cdot 1}{\frac{1}{\sqrt{2}}} = 2\sqrt{2} \)
Therefore, Equation of the tangent to the given curve at \( t = \frac{\pi}{4} \) is
\( y - 0 = 2\sqrt{2} \left( x - \frac{1}{\sqrt{2}} \right) \)
\( \implies y = 2\sqrt{2}x - 2 \)

Question. Find the values of \( x \) for which \( f(x) = [x(x - 2)]^2 \) is an increasing function. Also, find the points on the curve where tangent is parallel to x-axis.
Answer: Refer to answer 11.
For tangent parallel to x-axis, \( f'(x) = 0 \)
\( \implies 4x (x - 1) (x - 2) = 0 \implies x = 0, x = 1 \text{ and } x = 2 \)
when \( x = 0, y = 0 \); when \( x = 1, y = 1 \); when \( x = 2, y = 0 \)
The points on the curve at which the tangents are parallel to x-axis are \( (0, 0), (1, 1) \) and \( (2, 0) \).

Question. Find the points on the curve \( y = x^3 \) at which the slope of the tangent is equal to y-coordinate of the point.
Answer: We have, \( y = x^3 \) ...(i)
Differentiating (i), w.r.t. \( x \), we get
\( \frac{dy}{dx} = 3x^2 \) ...(ii)
Since it is given that slope of tangent is equal to the y-coordinate of the point.
Therefore, \( \frac{dy}{dx} = y \implies 3x^2 = y \) [using (ii)]
\( \implies 3x^2 = x^3 \) [using (i)]
\( \implies x^2(3 - x) = 0 \)
\( \implies x = 0 \text{ or } x = 3 \)
When \( x = 0 \), then from (i), \( y = 0 \)
When \( x = 3 \), then from (i), \( y = 3^3 = 27 \)
Therefore, The required points are \( (0, 0) \) and \( (3, 27) \).

Question. Find the equation of the tangent to the curve \( y = \frac{x-7}{(x-2)(x-3)} \) at the point, where it cuts the x-axis.
Answer: The equation of curve is \( y = \frac{x-7}{(x-2)(x-3)} \).
This cuts the x-axis at the point where \( y = 0 \).
Putting \( y = 0 \) in the equation, we get \( x = 7 \).
Thus the point of contact is \( (7, 0) \).
Now, \( y = \frac{x-7}{x^2 - 5x + 6} \)
\( \implies \frac{dy}{dx} = \frac{(x^2-5x+6)-(x-7)(2x-5)}{(x^2-5x+6)^2} \)
\( \left(\frac{dy}{dx}\right)_{(7,0)} = \frac{(7^2-5\times 7+6)-(7-7)(2\times 7-5)}{(7^2-5\times 7+6)^2} = \frac{1}{20} \)
The equation of tangent at \( (7, 0) \) is
\( (y - 0) = \frac{1}{20}(x - 7) \)
\( \implies 20y - x + 7 = 0 \)

Question. Find the equation of tangent to the curve given by \( x = a \sin^3 t, y = b \cos^3 t \) at a point where \( t = \frac{\pi}{4} \).
Answer: Refer to answer 36.

Question. Find the equation of tangent to the curve \( x = \sin 3t, y = \cos 3t \), at \( t = \frac{\pi}{4} \).
Answer: We have, \( x = \sin 3t \) and \( y = \cos 3t \)
\( \implies \frac{dx}{dt} = 3 \cos 3t \) and \( \frac{dy}{dt} = -3 \sin 3t \)
Therefore, \( \frac{dy}{dx} = \frac{-3 \sin 3t}{3 \cos 3t} \implies \left(\frac{dy}{dx}\right)_{t = \frac{\pi}{4}} = 1 \)
At \( t = \frac{\pi}{4} \), \( x = \sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}} \) and \( y = \cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}} \)
Equation of tangent at \( t = \frac{\pi}{4} \) is
\( \left( y + \frac{1}{\sqrt{2}} \right) = 1 \left( x - \frac{1}{\sqrt{2}} \right) \implies x - y = \sqrt{2} \)

Question. At what points will the tangent to the curve \( y = 2x^3 - 15x^2 + 36x - 21 \) be parallel to x-axis. Also, find the equations of tangents to the curve at those points.
Answer: The given curve is \( y = 2x^3 - 15x^2 + 36x - 21 \).
\( \implies \frac{dy}{dx} = 6x^2 - 30x + 36 \)
For tangent parallel to x-axis, \( \frac{dy}{dx} = 0 \)
Therefore, \( 6x^2 - 30x + 36 = 0 \)
\( \implies 6(x^2 - 5x + 6) = 0 \implies 6(x - 3) (x - 2) = 0 \)
\( \implies x = 3, x = 2 \)
When \( x = 3 \), \( y = 6 \) and when \( x = 2 \), \( y = 7 \)
Therefore, The points are \( (3, 6) \) and \( (2, 7) \).
Equation of tangent at \( (3, 6) \) is,
\( (y - 6) = 0 \cdot (x - 3) \implies y = 6 \)
Similarly, equation of tangent at \( (2, 7) \) is,
\( y - 7 = 0 \implies y = 7 \)

Question. Prove that the curve \( \left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 \) touches the straight line \( \frac{x}{a} + \frac{y}{b} = 2 \) at \( (a, b) \) for all values of \( n \in \mathbb{N} \).
Answer: The given curve is \( \left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 \).
Differentiating w.r.t. \( x \), we get
\( \frac{n}{a}\left(\frac{x}{a}\right)^{n-1} + \frac{n}{b}\left(\frac{y}{b}\right)^{n-1} \frac{dy}{dx} = 0 \)
\( \implies \frac{dy}{dx} = -\left( \frac{x}{y} \right)^{n-1} \cdot \frac{b^n}{a^n} \cdot \frac{a^{n-1}}{b^{n-1}} = -\left(\frac{x/a}{y/b}\right)^{n-1} \frac{b}{a} \)
Slope of tangent at \( (a, b) \) is,
\( \left(\frac{dy}{dx}\right)_{(a,b)} = -\left(\frac{a/a}{b/b}\right)^{n-1} \frac{b}{a} = -\frac{b}{a} = m_1 \text{ (say)} \)
The equation of straight line is \( \frac{x}{a} + \frac{y}{b} = 2 \).
Slope of line \( = -\frac{b}{a} \)
Also point \( (a, b) \) satisfy the equation of straight line. Therefore, straight line touches the curve \( \left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 \).

Question. Find the equation of tangents to the curve \( y = \cos(x + y), -2\pi \le x \le 2\pi \) that are parallel to the line \( x + 2y = 0 \).
Answer: Let the point of contact of one of the tangents be \( (x_1, y_1) \).
Then \( (x_1, y_1) \) lies on \( y = \cos (x + y) \)
Therefore, \( y_1 = \cos(x_1 + y_1) \) ...(i)
Since the tangents are parallel to the line \( x + 2y = 0 \). Therefore,
Slope of tangent at \( (x_1, y_1) \) = slope of line \( x + 2y = 0 \)
\[ \implies \left(\frac{dy}{dx}\right)_{(x_1, y_1)} = -\frac{1}{2} \]
Since, the equation of curve is \( y = \cos(x + y) \)
Differentiating with respect to \( x \), we get
\( \frac{dy}{dx} = -\sin(x + y)\left(1 + \frac{dy}{dx}\right) \)
\[ \implies \left(\frac{dy}{dx}\right)_{(x_1, y_1)} = -\sin(x_1 + y_1)\left\{1 + \left(\frac{dy}{dx}\right)_{(x_1, y_1)}\right\} \]
\[ \implies -\frac{1}{2} = -\sin(x_1 + y_1)\left(1 - \frac{1}{2}\right) \]
\( \implies \sin(x_1 + y_1) = 1 \) ...(ii)
Squaring (i) and (ii), then adding,
\( \cos^2(x_1 + y_1) + \sin^2(x_1 + y_1) = y_1^2 + 1 \)
\( \implies y_1^2 + 1 = 1 \implies y_1 = 0 \)
Put \( y_1 = 0 \) in (i) and (ii),
\( \cos x_1 = 0 \) and \( \sin x_1 = 1 \)
\( \implies x_1 = \frac{\pi}{2}, -\frac{3\pi}{2} \)
Hence, the points of contact are \( \left(\frac{\pi}{2}, 0\right) \) and \( \left(-\frac{3\pi}{2}, 0\right) \).
The slope of the tangent is \( -\frac{1}{2} \).
Therefore, equation of tangents at \( \left(\frac{\pi}{2}, 0\right) \) and \( \left(-\frac{3\pi}{2}, 0\right) \) are
\( y - 0 = -\frac{1}{2}\left(x - \frac{\pi}{2}\right) \) and \( y - 0 = -\frac{1}{2}\left(x + \frac{3\pi}{2}\right) \)
or \( 2x + 4y - \pi = 0 \) and \( 2x + 4y + 3\pi = 0 \).

Question. Find the value of \( p \) for which the curves \( x^2 = 9p(9 - y) \) and \( x^2 = p(y + 1) \) cut each other at right angles.
Answer: The given curves are
\( x^2 = 9p(9 - y) \) ... (i)
\( x^2 = p(y + 1) \) ... (ii)
Clearly \( p \neq 0 \) (as \( p = 0 \implies \) both curves become same)
Solving (i) and (ii), we get
\( 9p(9 - y) = p(y + 1) \)
\( \implies 9(9 - y) = y + 1 \quad (\because p \neq 0) \)
\( \implies y = 8 \) and then \( x^2 = 9p \) ...(iii)
Diff. (i) w.r.t. \( y \), \( 2x \frac{dx}{dy} = 9p(-1) \)
\( \implies m_1 = \frac{dy}{dx} = -\frac{2x}{9p} \)
Differentiating (ii) w.r.t. \( y \), \( 2x \frac{dx}{dy} = p \implies m_2 = \frac{2x}{p} \)
For curves (i) and (ii) to cut each other at right angles,
\( m_1 \cdot m_2 = -1 \)
\( \implies \left(-\frac{2x}{9p}\right) \left(\frac{2x}{p}\right) = -1 \)
\( \implies 9p^2 = 4x^2 = 4 \cdot 9 p \) [Using (iii)]
\( \implies p(p - 4) = 0 \implies p = 4 \text{ as } p \neq 0 \)

Question. Find the equation of the tangent line to the curve \( y = x^2 - 2x + 7 \) which is (i) parallel to the line \( 2x - y + 9 = 0 \), (ii) perpendicular to the line \( 5y - 15x = 13 \).
Answer: The given curve is \( y = x^2 - 2x + 7 \) ...(i)
\( \implies \frac{dy}{dx} = 2x - 2 \)
= Slope of the tangent to (i) at \( (x, y) \) ...(ii)
(i) The tangent is parallel to the line \( 2x - y + 9 = 0 \)
Its slope = 2
Therefore, From (ii), \( 2x - 2 = 2 \implies x = 2 \)
From (i), \( y = 2^2 - 2 \times 2 + 7 = 7 \)
Therefore, Equation of the tangent to (i) at \( (2, 7) \) whose slope = 2, is
\( y - 7 = 2(x - 2) \implies 2x - y + 3 = 0 \)
(ii) The tangent is perpendicular to the line \( 5y - 15x = 13 \),
its slope \( = \frac{15}{5} = 3 \)
Therefore, From (ii), \( (2x - 2) \times 3 = -1 \)
\( \implies 2x - 2 = -\frac{1}{3} \implies x = \frac{5}{6} \)
From (i), \( y = \left(\frac{5}{6}\right)^2 - 2 \times \frac{5}{6} + 7 = \frac{217}{36} \)
Therefore, Equation of the tangent to (i) at \( \left(\frac{5}{6}, \frac{217}{36}\right) \) whose slope \( = -\frac{1}{3} \), is
\( y - \frac{217}{36} = -\frac{1}{3} \left( x - \frac{5}{6} \right) \)
\( \implies 36y - 217 = -12x + 10 \implies 12x + 36y = 227 \)

Question. Find the equation of the normal at a point on the curve \( x^2 = 4y \) which passes through the point \( (1, 2) \). Also find the equation of the corresponding tangent.
Answer: We have, \( x^2 = 4y \)
Differentiating w.r.t. \( x \), we get
\( 2x = 4 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2} \)
Let \( P(x_1, y_1) \) be the point on the given curve.
\( \left[\frac{dy}{dx}\right]_P = \frac{x_1}{2} \)
Equation of normal at \( P(x_1, y_1) \)
\( (y - y_1) = -\frac{2}{x_1}(x - x_1) \) ... (i)
Equation (i) passes through the point \( (1, 2) \)
\( (2 - y_1) = -\frac{2}{x_1}(1 - x_1) \implies 2 - y_1 = -\frac{2}{x_1} + 2 \)
\( \implies x_1 y_1 = 2 \) ...(ii)
Also \( P(x_1, y_1) \) lies on \( x^2 = 4y \)
Therefore, \( x_1^2 = 4y_1 \)
\( \implies x_1^2 = 4 \times \frac{2}{x_1} \) (Using (ii))
\( \implies x_1^3 = 8 \implies x_1 = 2 \),
From (ii), \( 2 \times y_1 = 2 \implies y_1 = 1 \)
Now, putting the value of \( x_1 \) and \( y_1 \) in equation (i), we get
\( (y - 1) = -\frac{2}{2}(x - 2) \)
\( \implies y - 1 = -x + 2 \implies x + y = 3 \)
Equation of tangent
\( (y - y_1) = \left[\frac{dy}{dx}\right]_P (x - x_1) \)
\( \implies y - 1 = \frac{2}{2}(x - 2) \implies x - y = 1 \)

Question. Find the equations of tangents to the curve \( 3x^2 - y^2 = 8 \), which pass through the point \( \left(\frac{4}{3}, 0\right) \).
Answer: We have, \( 3x^2 - y^2 = 8 \) ...(i)
Differentiating w.r.t. \( x \), we get
\( 6x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{3x}{y} \).
Let \( P(x_1, y_1) \) be a point of the curve
Therefore, \( 3x_1^2 - y_1^2 = 8 \) ...(ii)
and \( \left[\frac{dy}{dx}\right]_P = \frac{3x_1}{y_1} \)
Therefore, Equation of tangent to (i) at \( (x_1, y_1) \) is
\( y - y_1 = \left[\frac{dy}{dx}\right]_P (x - x_1) \)
\( \implies y - y_1 = \frac{3x_1}{y_1}(x - x_1) \)
The tangent is passes through \( \left(\frac{4}{3}, 0\right) \)
Therefore, \( 0 - y_1 = \frac{3x_1}{y_1}\left(\frac{4}{3} - x_1\right) \)
\( \implies -y_1^2 = 4x_1 - 3x_1^2 \)
\( \implies 4x_1 = 3x_1^2 - y_1^2 = 8 \) [Using (ii)]
\( \implies x_1 = 2 \)
Again from (ii), \( 3x_1^2 - y_1^2 = 8 \)
\( \implies y_1 = \pm 2 \)
Therefore, The two points on (i), at which tangents pass through \( \left(\frac{4}{3}, 0\right) \) are \( (2, 2) \) and \( (2, -2) \).
Therefore, Equations of tangents are
\( y - 2 = \frac{3 \times 2}{2}(x - 2) \implies y = 3x - 4 \)
and \( y + 2 = \frac{3 \times 2}{-2}(x - 2) \implies y = -3x + 4 \)

Question. For the curve \( y = 4x^3 - 2x^5 \), find all the points on the curve at which the tangent passes through the origin.
Answer: The given curve is
\( y = 4x^3 - 2x^5 \) ...(i)
\( \implies \frac{dy}{dx} = 12x^2 - 10x^4 \)
Let \( P(x_1, y_1) \) be a point on the curve
Therefore, \( y_1 = 4x_1^3 - 2x_1^5 \) ...(ii)
and \( \left[\frac{dy}{dx}\right]_P = 12x_1^2 - 10x_1^4 \)
Now tangent to (i) at \( P \) is
\( y - y_1 = (12x_1^2 - 10x_1^4) \cdot (x - x_1) \)
This will pass through \( (0, 0) \) when
\( 0 - y_1 = (12x_1^2 - 10x_1^4)(0 - x_1) \)
\( \implies y_1 = 12x_1^3 - 10x_1^5 \)
\( \implies 4x_1^3 - 2x_1^5 = 12x_1^3 - 10x_1^5 \) [Using (ii)]
\( \implies 8x_1^3 - 8x_1^5 = 0 \implies x_1^3(1 - x_1^2) = 0 \)
\( \implies x_1 = 0, 1, -1 \).
Now, \( x_1 = 0 \), from (i), \( y_1 = 0 \)
\( x_1 = 1 \), from (i), \( y_1 = 2 \)
\( x_1 = -1 \), from (i), \( y_1 = -2 \)
Therefore, Tangents to given curve will pass through the origin at the points \( (0, 0), (1, 2) \) and \( (-1, -2) \).

Question. Find the equation of the tangent to the curve \( y = \sqrt{3x - 2} \) which is parallel to the line \( 4x - 2y + 5 = 0 \).
Answer: The given curve is \( y = \sqrt{3x - 2} \) ...(i)
\( \implies \frac{dy}{dx} = \frac{1}{2}(3x - 2)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x - 2}} \)
We want the tangent to (i) that is parallel to the line \( 4x - 2y + 5 = 0 \) ...(ii)
Therefore, Slope of tangent = slope of line \( 4x - 2y + 5 = 0 \)
\( \implies \frac{3}{2\sqrt{3x - 2}} = -\frac{4}{-2} = 2 \)
\( \implies 3 = 4\sqrt{3x - 2} \implies 9 = 16(3x - 2) \)
\( \implies 3x - 2 = \frac{9}{16} \implies 3x = 2 + \frac{9}{16} \implies x = \frac{41}{48} \)
From (i),
\( y = \sqrt{3 \cdot \frac{41}{48} - 2} = \frac{3}{4} \)
Therefore, The equation of the tangent to (i) at \( \left(\frac{41}{48}, \frac{3}{4}\right) \) that have slope = 2 is
\( y - \frac{3}{4} = 2 \left( x - \frac{41}{48} \right) \implies 4y - 3 = 8x - \frac{41}{6} \)
\( \implies 8x - 4y - \frac{41}{6} + 3 = 0 \implies 8x - 4y - \frac{23}{6} = 0 \)

Question. Prove that the curve \( x = y^2 \) and \( xy = k \) cut at the right angles if \( 8k^2 = 1 \).
Answer: The given curves are
\( y^2 = x \) ... (i)
\( xy = k \) ...(ii)
Differentiating (i) w.r.t. \( x \), we get
\( 2y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y} = m_1 \text{ (say)} \)
Differentiating (ii) w.r.t. \( x \), we get
\( x \frac{dy}{dx} + 1 \cdot y = 0 \implies \frac{dy}{dx} = -\frac{y}{x} = m_2 \text{ (say)} \)
Now (i) and (ii) will intersect orthogonally if
\( m_1 \cdot m_2 = -1 \)
\( \implies \left(\frac{1}{2y}\right)\left(-\frac{y}{x}\right) = -1 \)
\( \implies 2x = 1 \implies x = \frac{1}{2} \)
From (ii), \( y = \frac{k}{x} = 2k \quad \left(\because x = \frac{1}{2}\right) \)
Substituting the values of \( x \) and \( y \) in (i), we get
\( (2k)^2 = \frac{1}{2} \implies 8k^2 = 1 \)
which is the required condition for the orthogonality of the curves (i) and (ii).

Question. Prove that all normals to the curve \( x = a \cos t + at \sin t, y = a \sin t - at \cos t \) are at a constant distance 'a' from the origin.
Answer: Here, \( x = a \cos t + at \sin t \)
\( y = a \sin t - at \cos t \)
Now, \( \frac{dx}{dt} = -a \sin t + a \sin t + at \cos t = at \cos t \)
and \( \frac{dy}{dt} = a \cos t - a[1 \cdot \cos t - t \sin t] = at \sin t \)
Therefore, Slope of the tangent \( = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \tan t \)
\( \implies \text{Slope of the normal} = -\frac{1}{\tan t} = -\frac{\cos t}{\sin t} \)
Therefore, Equation of the normal to the given curve
\( y - (a \sin t - at \cos t) = -\frac{\cos t}{\sin t}(x - a \cos t - at \sin t) \)
\( \implies y \sin t - a \sin^2 t + at \sin t \cos t = -x \cos t + a \cos^2 t + at \sin t \cos t \)
\( \implies x \cos t + y \sin t - a = 0 \)
Therefore, Perpendicular distance from \( (0, 0) \) on it
\[ = \frac{|0 \cdot \cos t + 0 \cdot \sin t - a|}{\sqrt{\cos^2 t + \sin^2 t}} = a \]

Question. Find the equations of the tangent and normal to the parabola \( y^2 = 4ax \) at the point \( (at^2, 2at) \).
Answer: The given parabola is \( y^2 = 4ax \) ...(i)
and the point is \( P(at^2, 2at) \).
Differentiating (i) w.r.t. \( x \), we get
\( 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} \) ...(ii)
Therefore, Slope of the tangent at \( P = \frac{2a}{2at} = \frac{1}{t} \)
Equation of the tangent to (i) at \( P \) is
\( y - 2at = \frac{1}{t}(x - at^2) \)
\( \implies t y - 2at^2 = x - at^2 \implies ty = x + at^2 \)
From (ii), \( -\frac{dx}{dy} = -\frac{y}{2a} \)
Therefore, Slope of the normal at \( P = -\frac{2at}{2a} = -t \)
Therefore, Equation of the normal to (i) at \( P \) is
\( y - 2at = -t(x - at^2) \)
\( \implies y + tx - 2at - at^3 = 0 \)