Tangent and Normal JEE Mathematics Worksheets Set 01

Question. Find the points on the curve \( y = x^3 \) at which at slope of the tangent is equal to the y-coordinates of the point.
Answer: Let the point be \( P(x_1, y_1) \).
\( y_1 = x_1^3 \)
\( \frac{dy}{dx} = 3x^2 \)
At point \( P \), \( \frac{dy}{dx} = 3x_1^2 \)
Given, \( 3x_1^2 = y_1 \)
\( \implies 3x_1^2 = x_1^3 \)
\( \implies x_1 = 0, 3 \)
If \( x_1 = 0 \), then \( y_1 = 0 \)
If \( x_1 = 3 \), then \( y_1 = 27 \)
Points are \( (0, 0) \) and \( (3, 27) \).

Question. Find the equation of tangent to the curve \( y = 1 + e^{-2x} \) where it cuts the line \( y = 2 \).
Answer: The point of intersection is given by \( y = 2 \).
\( 2 = 1 + e^{-2x} \)
\( \implies e^{-2x} = 1 \)
\( \implies -2x = 0 \)
\( \implies x_1 = 0, y_1 = 2 \)
Point is \( (0, 2) \).
\( \frac{dy}{dx} = -2e^{-2x} \)
At point \( (0, 2) \), \( \frac{dy}{dx} = -2 \).
Equation of tangent: \( y - 2 = -2(x - 0) \)
\( \implies y + 2x = 2 \).

Question. Find equation of tangent and normal to the curve \( x^3 + y^3 = 6xy \) at point \( (3, 3) \).
Answer: Differentiating \( x^3 + y^3 = 6xy \) with respect to \( x \):
\( 3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx} \)
At point \( (3, 3) \):
\( 3(3)^2 + 3(3)^2 \frac{dy}{dx} = 6(3) + 6(3) \frac{dy}{dx} \)
\( \implies 27 + 27 \frac{dy}{dx} = 18 + 18 \frac{dy}{dx} \)
\( \implies 9 \frac{dy}{dx} = -9 \)
\( \implies \frac{dy}{dx} = -1 \).
Equation of tangent: \( y - 3 = -1(x - 3) \)
\( \implies y - 3 = -x + 3 \)
\( \implies x + y = 6 \).
Equation of normal: slope of normal \( = 1 \).
\( y - 3 = 1(x - 3) \)
\( \implies y = x \).

Question. Find the equation of normal to the curve \( x^3 + y^3 = 8xy \) at point where it is meet by the curve \( y^2 = 4x \), other than origin.
Answer: Solving \( x^3 + y^3 = 8xy \) and \( y^2 = 4x \):
\( (\frac{y^2}{4})^3 + y^3 = 8(\frac{y^2}{4})y \)
\( \implies \frac{y^6}{64} + y^3 = 2y^3 \)
\( \implies \frac{y^6}{64} = y^3 \)
\( \implies y^3 = 64 \implies y = 4 \).
As \( y = 4 \), \( x = \frac{4^2}{4} = 4 \).
Point is \( (4, 4) \).
Differentiating \( x^3 + y^3 = 8xy \):
\( 3x^2 + 3y^2 y' = 8y + 8xy' \)
At point \( (4, 4) \):
\( 3(4)^2 + 3(4)^2 y' = 8(4) + 8(4) y' \)
\( \implies 48 + 48y' = 32 + 32y' \)
\( \implies 16y' = -16 \)
\( \implies y' = -1 \).
Slope of normal \( = 1 \).
Equation of normal: \( y - 4 = 1(x - 4) \)
\( \implies y = x \).

Question. 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 \( 2y - x + 1 = 0 \)
Answer: \( \frac{dy}{dx} = 2x - 2 \).
(a) Parallel to \( y = 2x + 9 \), slope \( = 2 \).
\( 2x_1 - 2 = 2 \implies x_1 = 2 \).
\( y_1 = (2)^2 - 2(2) + 7 = 7 \).
Equation of tangent: \( y - 7 = 2(x - 2) \)
\( \implies y - 2x - 3 = 0 \).
(b) Perpendicular to \( y = \frac{1}{2}x - \frac{1}{2} \), slope \( = -2 \).
\( 2x_1 - 2 = -2 \implies x_1 = 0 \).
\( y_1 = (0)^2 - 2(0) + 7 = 7 \).
Equation of tangent: \( y - 7 = -2(x - 0) \)
\( \implies 2x + y - 7 = 0 \).

Question. A particle moves along the curve \( 6y = x^3 + 2 \). Find the points on the curve at which the y coordinate is changing 8 times as fast as the x co-ordinate.
Answer: Curve is \( 6y = x^3 + 2 \).
Differentiating with respect to \( t \): \( 6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \).
Given \( \frac{dy}{dt} = 8 \frac{dx}{dt} \).
\( \implies 6(8 \frac{dx}{dt}) = 3x^2 \frac{dx}{dt} \)
\( \implies 48 = 3x^2 \implies x^2 = 16 \implies x = \pm 4 \).
If \( x = 4 \), then \( 6y = 4^3 + 2 = 66 \implies y = 11 \).
If \( x = -4 \), then \( 6y = (-4)^3 + 2 = -62 \implies y = -31/3 \).
Points are \( (4, 11) \) and \( (-4, -31/3) \).

Question. For the curve \( y = 4x^3 - 2x^5 \), find all the points at which the tangent passes through the origin.
Answer: Let the point of contact be \( P(x_1, y_1) \).
\( y_1 = 4x_1^3 - 2x_1^5 \)
\( \frac{dy}{dx} = 12x^2 - 10x^4 \). Slope at \( P \) is \( 12x_1^2 - 10x_1^4 \).
Equation of tangent: \( y - y_1 = (12x_1^2 - 10x_1^4)(x - x_1) \).
Since it passes through origin \( (0, 0) \):
\( -y_1 = (12x_1^2 - 10x_1^4)(-x_1) \)
\( y_1 = 12x_1^3 - 10x_1^5 \)
Substitute \( y_1 \): \( 4x_1^3 - 2x_1^5 = 12x_1^3 - 10x_1^5 \)
\( \implies 8x_1^5 - 8x_1^3 = 0 \)
\( \implies 8x_1^3(x_1^2 - 1) = 0 \implies x_1 = 0, \pm 1 \).
Points are \( (0, 0), (1, 2), (-1, -2) \).

Question. If the tangent at \( (1, 1) \) on \( y^2 = x(2 - x)^2 \) meets the curve again at \( P \), then find coordinates of \( P \).
Answer: \( y^2 = x(2 - x)^2 \).
Differentiating: \( 2y \frac{dy}{dx} = (2 - x)^2 + x \cdot 2(2 - x)(-1) = (2 - x)(2 - x - 2x) = (2 - x)(2 - 3x) \).
At \( (1, 1) \), \( 2(1) \frac{dy}{dx} = (2 - 1)(2 - 3) = -1 \implies \frac{dy}{dx} = -1/2 \).
Equation of tangent: \( y - 1 = -\frac{1}{2}(x - 1) \implies 2y - 2 = -x + 1 \implies x = 3 - 2y \).
Substituting in curve: \( y^2 = (3 - 2y)(2 - (3 - 2y))^2 = (3 - 2y)(2y - 1)^2 \)
\( \implies y^2 = (3 - 2y)(4y^2 - 4y + 1) \)
\( \implies y^2 = 12y^2 - 12y + 3 - 8y^3 + 8y^2 - 2y \)
\( \implies 8y^3 - 19y^2 + 14y - 3 = 0 \).
Since \( y = 1 \) is a root (point of tangency), we can divide by \( (y-1)^2 \):
\( (y-1)^2(8y-3) = 0 \implies y = 3/8 \).
Then \( x = 3 - 2(3/8) = 3 - 3/4 = 9/4 \).
Point \( P \) is \( (9/4, 3/8) \).

Question. Find the angle of intersection of the curve \( y = 2 \sin^2 x \) and \( y = \cos 2x \).
Answer: For intersection point: \( 2 \sin^2 x = \cos 2x = 1 - 2 \sin^2 x \)
\( \implies 4 \sin^2 x = 1 \implies \sin^2 x = 1/4 \implies \sin x = \pm 1/2 \).
For \( y = 2 \sin^2 x \), \( m_1 = \frac{dy}{dx} = 4 \sin x \cos x = 2 \sin 2x \).
For \( y = \cos 2x \), \( m_2 = \frac{dy}{dx} = -2 \sin 2x \).
If \( \sin x = 1/2 \), then \( x = \pi/6 \), \( \sin 2x = \sin(\pi/3) = \sqrt{3}/2 \).
\( m_1 = \sqrt{3} \), \( m_2 = -\sqrt{3} \).
\( \tan \theta = | \frac{m_1 - m_2}{1 + m_1 m_2} | = | \frac{\sqrt{3} - (-\sqrt{3})}{1 + (\sqrt{3})(-\sqrt{3})} | = | \frac{2 \sqrt{3}}{1 - 3} | = | \frac{2 \sqrt{3}}{-2} | = \sqrt{3} \).
\( \implies \theta = \pi/3 \).

Question. Show that subnormal at any point on the curve \( x^2 y^2 = a^2(x^2 - a^2) \) varies inversely as the cube of its abscissa.
Answer: Curve is \( y^2 = \frac{a^2(x^2 - a^2)}{x^2} = a^2 - \frac{a^4}{x^2} \).
Differentiating with respect to \( x \):
\( 2y \frac{dy}{dx} = \frac{2a^4}{x^3} \)
Length of subnormal \( SN = y \frac{dy}{dx} = \frac{a^4}{x^3} \).
Thus, \( SN \propto \frac{1}{x^3} \). Proved.

Question. If x and y are sides of two squares such that \( y = x - x^2 \). Find the rate of change of area of second square with respect to the first square when \( x = 1 \text{ cm} \).
Answer: Area of first square \( A_1 = x^2 \), Area of second square \( A_2 = y^2 \).
We need \( \frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx} \).
\( \frac{dA_1}{dx} = 2x \).
\( A_2 = (x - x^2)^2 \).
\( \frac{dA_2}{dx} = 2(x - x^2)(1 - 2x) \).
At \( x = 1 \), \( \frac{dA_1}{dx} = 2 \) and \( \frac{dA_2}{dx} = 2(0)(-1) = 0 \).
So, \( \frac{dA_2}{dA_1} = \frac{0}{2} = 0 \).

Question. The tangent to the graph of the function \( y = f(x) \) at the point with abscissa \( x = a \) forms with the x-axis an angle of \( \pi/3 \) and at the point with abscissa \( x = b \) at an angle of \( \pi/4 \), then find the value of the integral, \( \int_a^b f'(x) \cdot f''(x) dx \) [assume \( f''(x) \) to be continuous]
Answer: Given \( f'(a) = \tan(\pi/3) = \sqrt{3} \) and \( f'(b) = \tan(\pi/4) = 1 \).
Integral \( I = \int_a^b f'(x) \cdot f''(x) dx \).
Let \( f'(x) = t \), then \( f''(x) dx = dt \).
\( I = [ \frac{(f'(x))^2}{2} ]_a^b \)
\( I = \frac{1}{2} [ (f'(b))^2 - (f'(a))^2 ] \)
\( I = \frac{1}{2} [ 1^2 - (\sqrt{3})^2 ] = \frac{1}{2} (1 - 3) = -1 \).

Question. Find the abscissa of the point on the curve, \( xy = (c - x)^2 \) the normal at which cuts off numerically equal intercepts from the axes of co-ordinates.
Answer: If normal cuts equal intercepts, its slope is \( \pm 1 \).
Curve is \( xy = (c - x)^2 \).
Differentiating: \( x y' + y = 2(c - x)(-1) = 2(x - c) \).
\( y' = \frac{2(x - c) - y}{x} \).
For normal, \( -\frac{1}{y'} = \pm 1 \implies y' = \mp 1 \).
Consider \( y' = -1 \): \( \frac{2(x - c) - y}{x} = -1 \)
\( \implies 2x - 2c - y = -x \implies 3x - 2c = y \).
Substituting in curve: \( x(3x - 2c) = (c - x)^2 \)
\( \implies 3x^2 - 2cx = c^2 - 2cx + x^2 \)
\( \implies 2x^2 = c^2 \implies x = \pm \frac{c}{\sqrt{2}} \).

Question. If the relation between subnormal SN and subtangent ST at any point S on the curve \( by^2 = (x + a)^3 \) is \( p(SN) = q(ST)^2 \), then find value of \( \frac{p}{q} \) in terms of b and a.
Answer: \( by^2 = (x+a)^3 \implies 2by y' = 3(x+a)^2 \implies y' = \frac{3(x+a)^2}{2by} \).
\( SN = y y' = \frac{3(x+a)^2}{2b} \).
\( ST = y/y' = \frac{2by^2}{3(x+a)^2} = \frac{2(x+a)^3}{3(x+a)^2} = \frac{2(x+a)}{3} \).
\( p(SN) = q(ST)^2 \implies p \frac{3(x+a)^2}{2b} = q \frac{4(x+a)^2}{9} \)
\( \implies \frac{p}{q} = \frac{4/9}{3/2b} = \frac{8b}{27} \).

Question. In the curve \( x = a(\cos t + \ln \tan(t/2)) \), \( y = a \sin t \), show that the portion of the tangent between the point of contact and the x-axis is of constant length.
Answer: \( \frac{dx}{dt} = a(-\sin t + \frac{1}{\sin t}) = \frac{a \cos^2 t}{\sin t} \).
\( \frac{dy}{dt} = a \cos t \).
\( \frac{dy}{dx} = \frac{a \cos t}{a \cos^2 t / \sin t} = \tan t \).
Tangent equation: \( Y - a \sin t = \tan t(X - x_c) \).
For x-axis intersection point \( Q \), \( Y = 0 \):
\( -a \sin t = \tan t(X - x_c) \)
\( \implies -a \cos t = X - x_c \).
Point \( Q = (x_c - a \cos t, 0) \).
Length \( PQ = \sqrt{(x_c - (x_c - a \cos t))^2 + (a \sin t - 0)^2} = \sqrt{a^2 \cos^2 t + a^2 \sin^2 t} = a \).
Constant length.

Question. For the curve \( 6y = x^3 + 2 \), find all the points at which the tangent has a given slope.
Answer: Given \( \frac{dy}{dt} = 8 \frac{dx}{dt} \). This implies \( \frac{dy}{dx} = 8 \).
\( 6 \frac{dy}{dx} = 3x^2 \implies 6(8) = 3x^2 \)
\( \implies 48 = 3x^2 \implies x^2 = 16 \implies x = \pm 4 \).
At \( x = 4 \), \( y = 11 \).
At \( x = -4 \), \( y = -31/3 \).
Points are \( (4, 11) \) and \( (-4, -31/3) \).

Question. Find the point of intersection of the tangents drawn to the curve \( x^2y = 1 - y \) at the points where it is intersected by the curve \( xy = 1 - y \).
Answer: \( x^2y = 1 - y \)
\( xy = 1 - y \)
\( \implies x^2y = xy \)
\( \implies x^2y - xy = 0 \)
\( \implies xy(x - 1) = 0 \)
\( xy = 0 \) & \( x = 1 \)
\( \implies y = 1/2 \)
\( x = 0 \)
\( \implies y = 1 \) POI \( P(1, 1/2) \)
\( y = 0 \), \( 0 = 1 \) not possible \( Q(0, 1) \)
\( x^2y = 1 - y \)
\( 2xy + x^2y' = -y' \)
\( y' = \frac{-2xy}{1 + x^2}|_p = \frac{-2(1)(1/2)}{1 + 1} = -\frac{1}{2} \)
\( y'|_Q = 0 \)
at P tangent: \( y - \frac{1}{2} = -\frac{1}{2}(x - 1) \) ....(1)
at Q tangent: \( y - 1 = 0 \)
\( \implies y = 1 \) ....(2)
Intersection of (1) & (2):
\( \frac{1}{2} = -\frac{1}{2}(x - 1) \)
\( x - 1 = -1 \)
\( \implies x = 0 \)
POI (0, 1)

Question. Find all the lines that pass through the point (1, 1) and are tangent to the curve represented parametrically as \( x = 2t - t^2 \) and \( y = t + t^2 \).
Answer: \( \frac{dy}{dx} = \frac{2t + 1}{2 - 2t} \)
equation of tangent:
\( y - (t + t^2) = \frac{2t + 1}{2 - 2t}(x - (2t - t^2)) \)
at (1, 1)
\( \implies 2t^2 + 2t - 2 = 2t^2 - t - 1 \)
so ; \( t = 1/3 \); \( m = 5/4 \)
\( t = 1 \); \( m \to \infty \)

Question. A function is defined parametrically by the equations \( f(t) = x = \begin{cases} 2t + t^2 \sin \frac{1}{t} & \text{if } t \neq 0 \\ 0 & \text{if } t = 0 \end{cases} \) and \( g(t) = y = \begin{cases} \frac{1}{t} \sin^2 t & \text{if } t \neq 0 \\ 0 & \text{if } t = 0 \end{cases} \). Find the equation of the tangent and normal at the point for \( t = 0 \) if exist.
Answer: \( y'(0) = \text{Lim}_{h \to 0} \frac{1/h \sin^2 h - 0}{h} = 1 \)
\( x'(0) = \text{Lim}_{h \to 0} \frac{2h + h^2 \sin 1/h - 0}{h} = 2 \)
so \( m = 1/2 \)
T : \( y = 1/2x \)
N : \( y = -2x \)

Question. Find all the 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: \( y = \cos (x + y) \)
\( y' = -\sin (x + y)(1 + y') \)
\( y' = -\frac{\sin (x + y)}{1 + \sin (x + y)} = -\frac{1}{2} \)
\( \implies \sin (x + y) = 1 \)
\( \cos (x + y) = 0 \)
\( y_1 = \cos (x_1 + y_1) = 0 \)
\( \sin (x_1 + y_1) = 1 \)
\( \implies \sin x_1 = 1 \)
\( x_1 = \frac{\pi}{2}, -\frac{3\pi}{2} \)
Point \( \left( \frac{\pi}{2}, 0 \right) \) and \( \left( -\frac{3\pi}{2}, 0 \right) \)
This two points satisfies

Question. Show that the normals to the curve \( x = a(\cos t + t \sin t) \); \( y = a(\sin t - t \cos t) \) are tangent lines to the circle \( x^2 + y^2 = a^2 \).
Answer: \( \frac{dy}{dx} = \frac{a(\cos t - \cos t + t \sin t)}{a(-\sin t + \sin t + t \cos t)} = \tan t \)
equation normal
\( y - a(\sin t - t \cos t) = -\frac{\cos x}{\sin x}(x - a \cos t - a t \sin t) \)
\( \implies y \sin t - a \sin^2 t + at \sin t \cos t = -x \cos t + a \cos^2 t + a t \sin t \cos t \)
\( \implies x \cos t + y \sin t = a \)
tangent to given circle.

Question. If the tangent at the point \( (x_1, y_1) \) to the curve \( x^3 + y^3 = a^3 \) meets the curve again in \( (x_2, y_2) \) then show that \( \frac{x_2}{x_1} + \frac{y_2}{y_1} = -1 \).
Answer: \( x_1^3 + y_1^3 = a^3 \) & \( x_2^3 + y_2^3 = a^3 \) subtract
\( \implies (x_2^3 - x_1^3) + (y_2^3 - y_1^3) = 0 \) .....(i)
\( \frac{dy}{dx}|_{(x_1, y_1)} = -\frac{x_1^2}{y_1^2} \) .....(ii)
also slope = \( \frac{y_2 - y_1}{x_2 - x_1} = -\left( \frac{x_2^2 + x_1^2 + x_1x_2}{y_2^2 + y_1^2 + y_1y_2} \right) \) ..(iii)
from (ii) & (iii)
\( \implies -\frac{x_1^2}{y_1^2} = -\left( \frac{x_2^2 + x_1^2 + x_1x_2}{y_2^2 + y_1^2 + y_1y_2} \right) \)
on solving : \( \frac{x_2}{x_1} + \frac{y_2}{y_1} = -1 \)

Question. The tangent at a variable point P of the curve \( y = x^2 - x^3 \) meets it again at Q. Show that the locus of the middle point of PQ is \( y = 1 - 9x + 28x^2 - 28x^3 \).
Answer: \( y = x^2 - x^3 \)
Let \( P(t_1, t_1^2 - t_1^3) \) & \( Q(t_2, t_2^2 - t_2^3) \)
\( \frac{dy}{dx} = 2x - 3x^2|_p = 2t_1 - 3t_1^2 \)
\( 2t_1 - 3t_1^2 = M_{PQ} \)
\( 2t_1 - 3t_1^2 = \frac{t_2^2 - t_2^3 - t_1^2 + t_1^3}{t_2 - t_1} \)
\( 2t_1 - 3t_1^2 = (t_2 + t_1) - (t_2^2 + t_1^2 + t_1 t_2) \)
\( \implies t_2 + 2t_1 - 1 = 0 \) ......(1)
Now \( 2h = t_1 + t_2 \) ......(2)
\( 2k = t_1^2 - t_1^3 + t_2^2 - t_2^3 \) ......(3)
from (1) & (2)
\( t_1 = 1 - 2h \)
& \( t_2 = 3 - 4h \)
put the values is equation (3)
\( k = 1 - 9h + 28h^2 - 28h^3 \)
Locus \( \implies y = 1 - 9x + 28x^2 - 28x^3 \)

Question. Show that the condition that the curves \( x^{2/3} + y^{2/3} = c^{2/3} \) & \( (x^2/a^2) + (y^2/b^2) = 1 \) may touch if \( c = a + b \).
Answer: Let the point \( P(c \cos^3 \theta, c \sin^3 \theta) \) lie on the curve \( x^{2/3} + y^{2/3} = c^{2/3} \)
\( \frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}} = -\tan \theta \)
tangent at point P
\( y - c \sin^3 \theta = -\tan \theta (x - c \cos^3 \theta) \)
\( y = -\tan \theta x + c \tan \theta \cos^3 \theta + c \sin^3 \theta \)
Now apply condition of tangency of ellipse
\( c^2 = b^2 + a^2 m^2 (c \tan \theta \cos^3 \theta + c \sin^3 \theta)^2 = b^2 + a^2 \tan^2 \theta \)
\( \implies c = a + b \)

Question. A curve is given by the equations \( x = at^2 \) & \( y = at^3 \). A variable pair of perpendicular lines through the origin 'O' meet the curve at P & Q. Show that the locus of the point of intersection of the tangents at P & Q is \( 4y^2 = 3ax - a^2 \).
Answer: for P & Q : \( t_1 t_2 = -1 \)
\( P = (at^2, at^3), Q = \left( \frac{a}{t^2}, -\frac{a}{t^3} \right) \)
\( T_1 : y - at^3 = \frac{3t}{2}(x - at^2) \)
\( T_2 : y + \frac{a}{t^3} = -\frac{3t}{2}\left(x - \frac{a}{t^2}\right) \)
solve these equation for point of intersection
Locus: \( 4y^2 = 3ax - a^2 \)

Question. A and B are points of the parabola \( y = x^2 \). The tangents at A and B meet at C. The median of the triangle ABC from C has length 'm' units. Find the area of the triangle in terms of 'm'.
Answer: Let point A & B are \( (t_1, t_1^2), (t_2, t_2^2) \)
\( T_1 : y = 2t_1x - t_1^2 \) .........(i)
\( T_2 : y = 2t_2x - t_2^2 \) .........(ii)
On solving : \( C = \left( \frac{t_1 + t_2}{2}, t_1 t_2 \right) \)
Mid point of A & B = \( \left( \frac{t_1 + t_2}{2}, \frac{t_1^2 + t_2^2}{2} \right) \)
Find the length of median & also perpendicular distance from A to median.
Now area = 2.(area of half \( \Delta \)).
Area = \( m^2 \)

Question. (a) Find the value of n so that the subnormal at any point on the curve \( xy^n = a^{n+1} \) may be constant. (b) Show that in the curve \( y = a \cdot \ln (x^2 - a^2) \), sum of the length of tangent & subtangent varies as the product of the coordinates of the point of contact.
Answer: (a) Take log & \( \frac{dy}{dx}|_{(x_1, y_1)} = -\frac{y_1}{nx_1} \)
\( L_{SN} = y_1 \left( -\frac{y_1}{nx_1} \right) = -\frac{y_1^2}{nx_1} = -\frac{y_1^{n+2}}{n \cdot a^{n+1}} \)
\( n + 2 = 0 \)
\( \implies n = -2 \)
(b) \( m = \frac{dy}{dx}|_{(x_1, y_1)} = \frac{2ax_1}{x_1^2 - a^2} \)
\( L_T + L_{ST} = y_1 \sqrt{1 + \frac{1}{m^2}} + \frac{y_1}{m} \)
\( = y_1 \left[ \frac{\sqrt{x_1^4 + a^4 + 2a^2x_1^2}}{2ax_1} + \frac{x_1^2 - a^2}{2ax_1} \right] \)
\( = \frac{y_1}{2ax_1}(x_1^2 + a^2 + x_1^2 - a^2) = \frac{x_1 y_1}{2a} \)

Question. If the two curves \( C_1 : x = y^2 \) and \( C_2 : xy = k \) cut at right angles find the value of k.
Answer: \( \left( \frac{dy}{dx} \right)_{C_1} \cdot \left( \frac{dy}{dx} \right)_{C_2} = -1 \)
\( \left( \frac{1}{2y_1} \right) \cdot \left( -\frac{y_1}{x_1} \right) = -1 \)
\( \implies x_1 = 1/2 \)
from \( I^{st} \) curve : \( y_1 = \pm \frac{1}{\sqrt{2}} \)
\( K = \pm \frac{1}{2\sqrt{2}} \)

Question. An open can of oil is accidentally dropped into a lake; assume the oil spreads over the surface as a circular disc of uniform thickness whose radius increases steadily at the rate of 10 cm/sec. At the moment when the radius is 1 meter, the thickness of the oil slick is decreasing at the rate of 4 mm/sec, how fast is it decreasing when the radius is 2 meters.
Answer: \( \frac{dr}{dt} = 10 \text{ cm/sec, } \frac{dh}{dt} = -0.4 \text{ cm/sec} \)
\( v = \pi r^2 h, \frac{dv}{dt} = 2\pi rh \cdot \frac{dr}{dt} + \pi r^2 \frac{dh}{dt} = 0 \)
\( \implies h = 2 \text{ cm } (r = 100) \)
\( r_1^2 h_1 = r_2^2 h_2 \)
\( \implies (100)^2 \cdot 2 = (200)^2 \cdot h_2 \)
\( h_2 = 1/2 \text{ cm} \)
\( \frac{dv}{dt} = \pi r^2 \cdot \frac{dh}{dt} + 2\pi rh \cdot \frac{dr}{dt} = 0 \)
\( \implies 200 \cdot \frac{dh}{dt} = -2 \cdot \frac{1}{2} \cdot 10 \)
\( \implies \frac{dh}{dh} = -\frac{1}{20} \text{ cm/sec} \)

Question. A circular ink blot grows at the rate of 2 cm\(^2\) per second. Find the rate at which the radius is increasing after \( 2 \frac{6}{11} \) seconds. (Use \( \pi = \frac{22}{7} \))
Answer: \( \pi r^2 = A \)
\( \implies r = \sqrt{\frac{A}{\pi}} \)
\( \frac{dr}{dA} = \frac{1}{2\sqrt{A\pi}} \) & \( A = 2 \times \frac{28}{\pi} \)
\( \frac{dr}{dA} = \frac{1}{2\sqrt{2 \times \frac{28}{11} \times \frac{22}{7}}} = \frac{1}{8} \)
Now \( \frac{dr}{dt} = \frac{dr}{dA} \times \frac{dA}{dt} = \frac{1}{8} \times 2 = \frac{1}{4} \text{ cm/sec} \)

Question. Water is flowing out at the rate of 6 m\(^3\)/min from a reservoir shaped like a hemispherical bowl of radius R = 13 m. The volume of water in the hemispherical bowl is given by \( V = \frac{\pi}{3} \cdot y^2 (3R - y) \) when the water is y meter deep. Find (a) At what rate is the water level changing when the water is 8 m deep. (b) At what rate is the radius of the water surface changing when the water is 8 m deep.
Answer: \( \frac{dv}{dt} = \pi R \cdot 2y \cdot \frac{dy}{dt} - \pi y^2 \cdot \frac{dy}{dt} \)

Question. At time t > 0, the volume of sphere is increasing at a rate proportional to the reciprocal of its radius. At t = 0, the radius of the sphere is 1 unit and at t = 15 the radius is 2 units. (a) Find the radius of the sphere as a function of time t. (b) At what time t will the volume of the sphere be 27 times its volume at t = 0.
Answer: \( \frac{dv}{dt} \times \frac{1}{r} \)
\( \implies \frac{dv}{dt} = \frac{c}{r} \)
\( \implies \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} = \frac{c}{r} \)
\( \implies 4\pi \int r^3 \cdot dr = c \int dt \)
\( \implies \pi r^4 = ct + k \)
\( \implies r^4 = t + 1 \)
\( \implies r = (1 + t)^{1/4} \)

Question. Find the equation of the straight line which is tangent at one point and normal at another point of the curve, \( x = 3t^2, y = 2t^3 \). 
Answer: \( y = 2t^3 ; x = 3t^2 \)
\( \frac{dy}{dt} = 6t^2, \frac{dx}{dt} = 6t \)
\( \frac{dy}{dx} = \frac{6t^2}{6t} = t \)
\( \left. \frac{dy}{dx} \right|_P = t_1 \)
\( \left. \frac{dy}{dx} \right|_Q = t_2 \)
\( t_1 t_2 = -1 \) ......(1)
slope of \( PQ = \frac{2t_2^3 - 2t_1^3}{3t_2^2 - 3t_1^2} = t_1 \)
\( \implies \frac{2(t_2^2 + t_1t_2 + t_1^2)}{3(t_2 + t_1)} = t_1 \)
\( \implies 2t_2^2 + 2t_1t_2 + 2t_1^2 = 3t_1(t_2 + t_1) \)
\( \implies (t_1 + 2t_2) (t_1 - t_2) = 0 \)
\( t_1 \neq t_2 ; t_1 = -2t_2 \) ......(2)
from (1) & (2)
\( t_1^2 = 2 \)
\( \implies t_1 = \pm \sqrt{2} \)
Tangent at P
\( y - 2t_1^3 = t_1 (x - 3t_1^2) \)
put \( t_1 \) values
\( \sqrt{2} x + y - 2\sqrt{2} = 0 \) or \( \sqrt{2} x - y - 2\sqrt{2} = 0 \)

Question. If the normal to the curve, \( y = f(x) \) at the point (3, 4) makes an angle \( 3\pi/4 \) with the positive x-axis. The \( f'(3) = \) 
(a) -1
(b) -3/4
(c) 4/3
(d) 1
Answer: (d) 1

Question. The point(s) on the curve \( y^3 + 3x^2 = 12y \) where the tangent is vertical, is (are) 
(a) \( \left( \pm \frac{4}{\sqrt{3}}, -2 \right) \)
(b) \( \left( \pm \sqrt{\frac{11}{3}}, 1 \right) \)
(c) (0, 0)
(d) \( \left( \pm \frac{4}{\sqrt{3}}, 2 \right) \)
Answer: (d) \( \left( \pm \frac{4}{\sqrt{3}}, 2 \right) \)

Question. Tangent to the curve \( y = x^2 + 6 \) at a point P(1, 7) touches the circle \( x^2 + y^2 + 16x + 12y + c = 0 \) at a point Q. Then the coordinates of Q are 
(a) (-6, -11)
(b) (-9, -13)
(c) (-10, -15)
(d) (-6, -7)
Answer: (d) (-6, -7)

Question. The tangent to the curve \( y = e^x \) drawn at the point \( (c, e^c) \) intersects the line joining the points \( (c - 1, e^{c-1}) \) and \( (c + 1, e^{c+1}) \) 
(a) on the left of x = c
(b) on the right of x = c
(c) at no point
(d) at all points
Answer: (a) on the left of x = c