JEE Mathematics Tangent and Normal MCQs Set 01

Question. The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle \(x^2 + y^2 = 4\) at \((1, \sqrt{3})\) is
(a) \(3\sqrt{3}\) sq. units
(b) \(2\sqrt{3}\) sq. units
(c) \(4\sqrt{3}\) sq. units
(d) \(\sqrt{3}\) sq. units
Answer: (b) \(2\sqrt{3}\) sq. units

Question. Equation of the normal to the curve \(y = -\sqrt{x} + 2\) at the point of its intersection with the curve \(y = \tan(\tan^{-1} x)\) is
(a) \(2x - y - 1 = 0\)
(b) \(2x - y + 1 = 0\)
(c) \(2x + y - 3 = 0\)
(d) None of the options
Answer: (a) \(2x - y - 1 = 0\)

Question. The abscissa of the point on the curve \(ay^2 = x^3\), the normal at which cuts off equal intercepts from the coordinate axes is
(a) \(\frac{2a}{9}\)
(b) \(\frac{4a}{9}\)
(c) \(-\frac{4a}{9}\)
(d) \(-\frac{2a}{9}\)
Answer: (b) \(\frac{4a}{9}\)

Question. If the tangent to the curve \(x = a (\theta + \sin \theta)\), \(y = a (1 + \cos \theta)\) at \(\theta = \frac{\pi}{3}\) makes an angle \(\alpha\) (\(0 \leq \alpha < \pi\)) with x-axis, then \(\alpha\) equals
(a) \(\frac{\pi}{3}\)
(b) \(\frac{2\pi}{3}\)
(c) \(\frac{\pi}{6}\)
(d) \(\frac{5\pi}{6}\)
Answer: (d) \(\frac{5\pi}{6}\)

Question. The x-intercept of the tangent at any arbitrary point of the curve \(\frac{a}{x^2} + \frac{b}{y^2} = 1\) is proportional to
(a) square of the abscissa of the point of tangency
(b) square root of the abscissa of the point of tangency
(c) cube of the abscissa of the point of tangency
(d) cube root of the abscissa of the point of tangency.
Answer: (c) cube of the abscissa of the point of tangency

Question. If curve \(y = 1 - ax^2\) and \(y = x^2\) intersect orthogonally then the value of a is
(a) 1/2
(b) 1/3
(c) 2
(d) 3
Answer: (b) 1/3

Question. The coordinates of the point of the parabola \(y^2 = 8x\), which is at minimum distance from the circle \(x^2 + (y + 6)^2 = 1\) are
(a) \((2, -4)\)
(b) \((18, -12)\)
(c) \((2, 4)\)
(d) None of the options
Answer: (a) \((2, -4)\)

Question. The length of the subtangent to the curve \(\sqrt{x} + \sqrt{y} = 3\) at the point (4, 1) is
(a) 2
(b) 1/2
(c) 3
(d) 4
Answer: (a) 2

Question. For a curve \(\frac{(\text{length of normal})^2}{(\text{length of tangent})^2}\) is equal to
(a) (subnormal) / (subtangent)
(b) (subtangent) / (subnormal)
(c) (subnormal) / (subtangent)\(^2\)
(d) None of the options
Answer: (a) (subnormal) / (subtangent)

Question. Water is poured into an inverted conical vessel of which the radius of the base is 2m and height 4m, at the rate of 77 litre/minute. The rate at which the water level is rising at the instant when the depth is 70 cm is: (use \(\pi = 22/7\))
(a) 10 cm/min
(b) 20 cm/min
(c) 40 cm/min
(d) None of the options
Answer: (b) 20 cm/min

Question. If the tangent at each point of the curve \(y = \frac{2}{3}x^3 - 2ax^2 + 2x + 5\) makes an acute angle with the positive direction of x-axis, then
(a) \(a \geq 1\)
(b) \(-1 \leq a \leq 1\)
(c) \(a \leq -1\)
(d) None of the options
Answer: (b) \(-1 \leq a \leq 1\)

Question. The line \(\frac{x}{a} + \frac{y}{b} = 1\) touches the curve \(y = b e^{-x/a}\) at the point
(a) \((-a, be)\)
(b) \(\left(-a, \frac{b}{e}\right)\)
(c) \(\left(a, \frac{b}{e}\right)\)
(d) \((0, b)\)
Answer: (d) \((0, b)\)

Question. All points on the curve \(y^2 = 4a \left( x + a \sin \frac{x}{a} \right)\) at which the tangents are parallel to the axis of x, lie on a
(a) circle
(b) parabola
(c) line
(d) None of the options
Answer: (b) parabola

Question. A curve is represented by the equations, \(x = \sec^2 t\) and \(y = \cot t\) where t is a parameter. If the tangent at the point P on the curve where \(t = \pi/4\) meets the curve again at the point Q then |PQ| is equal to
(a) \(\frac{5\sqrt{3}}{2}\)
(b) \(\frac{5\sqrt{5}}{2}\)
(c) \(\frac{2\sqrt{5}}{3}\)
(d) \(\frac{3\sqrt{5}}{2}\)
Answer: (d) \(\frac{3\sqrt{5}}{2}\)

Question. If the subnormal at any point on \(y = a^{1-n} x^n\) is of constant length, then the value of n is
(a) 1
(b) 1/2
(c) 2
(d) -2
Answer: (b) 1/2

Question. The curves \(x^3 + p x y^2 = -2\) and \(3 x^2 y - y^3 = 2\) are orthogonal for
(a) p = 3
(b) p = -3
(c) no value of p
(d) p = \(\pm 3\)
Answer: (b) p = -3

Question. If curves \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and \(xy = c^2\) intersect orthogonally, then
(a) a + b = 0
(b) \(a^2 = b^2\)
(c) a + b = c
(d) None of the options
Answer: (b) \(a^2 = b^2\)

Question. The ordinate of \(y = (a/2) (e^{x/a} + e^{-x/a})\) is the geometric mean of the length of the normal and the quantity
(a) a/2
(b) a
(c) e
(d) None of the options
Answer: (b) a

Question. Angle between the tangents to the curve \(y = x^2 - 5x + 6\) at the points (2, 0) and (3, 0) is
(a) \(\pi/2\)
(b) \(\pi/6\)
(c) \(\pi/4\)
(d) \(\pi/3\)
Answer: (a) \(\pi/2\)

Question. If the tangent at P of the curve \(y^2 = x^3\) intersects the curve again at Q and the straight lines OP, OQ make angles \(\alpha, \beta\) with the x-axis, where 'O' is the origin, then \(\tan \alpha / \tan \beta\) has the value equal to
(a) -1
(b) -2
(c) 2
(d) \(\sqrt{2}\)
Answer: (b) -2

Question. Water is being poured on to a cylindrical vessel at the rate of 1 m\(^3\)/min. If the vessel has a circular base of radius 3m, the rate at which the level of water is rising in the vessel is
(a) \(1/9 \pi\) m/min
(b) \(0 \pi\) m/min
(c) \(1/3 \pi\) m/min
(d) \(3 \pi\) m/min
Answer: (a) \(1/9 \pi\) m/min

Question. Find the number of points on the curve \(x^2 + y^2 - 2x - 3 = 0\) at which the tangents are parallel to the x-axis.
(a) 1
(b) 2
(c) 3
(d) None of the options
Answer: (b) 2

Question. If at any point on a curve the subtangent and subnormal are equal, then the tangent is equal to
(a) ordinate
(b) \(\sqrt{2}\) ordinate
(c) \(\sqrt{2(\text{ordinate})}\)
(d) None of the options
Answer: (b) \(\sqrt{2}\) ordinate

Question. The length of the normal to the curve \(x = a(\theta + \sin \theta)\), \(y = a (1 - \cos \theta)\), at \(\theta = \frac{\pi}{2}\) is
(a) 2a
(b) \(a\sqrt{2}\)
(c) a/2
(d) \(a/\sqrt{2}\)
Answer: (b) \(a\sqrt{2}\)

Question. The number of values of c such that the straight line \(3x + 4y = c\) touches the curve \(\frac{x^4}{2} = x + y\) is
(a) 0
(b) 1
(c) 2
(d) 4
Answer: (b) 1

Question. The beds of two rivers (within a certain region) are a parabola \(y = x^2\) and a straight line \(y = x - 2\). These rivers are to be connected by a straight canal. The co-ordinates of the ends of the shortest canal can be
(a) \(\left(\frac{1}{2}, \frac{1}{4}\right)\) and \(\left(-\frac{11}{8}, \frac{5}{8}\right)\)
(b) \(\left(\frac{1}{2}, \frac{1}{4}\right)\) and \(\left(\frac{11}{8}, -\frac{5}{8}\right)\)
(c) (0, 0) and (1, -1)
(d) None of the options
Answer: (b) \(\left(\frac{1}{2}, \frac{1}{4}\right)\) and \(\left(\frac{11}{8}, -\frac{5}{8}\right)\)

Question. The points(s) of intersection of the tangents drawn to the curve \(x^2y = 1 - y\) at the points where it is intersected by the curve \(xy = 1 - y\) is/are given by
(a) (0, -1)
(b) (0, 1)
(c) (1, 1)
(d) None of the options
Answer: (b) (0, 1)

Question. If the area of the triangle included between the axes and any tangent to the curve \(x^n y = a^n\) is constant, then n is equal to
(a) 1
(b) 2
(c) 3/2
(d) 1/2
Answer: (a) 1

Question. At (0, 0), the curve \(y^2 = x^3 + x^2\)
(a) touches X-axis
(b) bisects the angle between the axes
(c) makes an angle of \(60^\circ\) with OX
(d) None of the options
Answer: (b) bisects the angle between the axes

Question. For the curve \(x = t^2 - 1\), \(y = t^2 - t\), the tangent line is perpendicular to x-axis where
(a) t = 0
(b) \(t = \infty\)
(c) \(t = \frac{1}{\sqrt{3}}\)
(d) \(t = -\frac{1}{\sqrt{3}}\)
Answer: (a) t = 0

Question. If tangent at point \( (1, 2) \) on the curve \( y = ax^2 + bx + \frac{7}{2} \) be parallel to normal at \( (-2, 2) \) on the curve \( y = x^2 + 6x + 10 \), then
(a) \( a = 1 \)
(b) \( a = -1 \)
(c) \( b = -5/2 \)
(d) \( b = 5/2 \)
Answer: (a) \( a = 1 \), (c) \( b = -5/2 \)
Solution:
\( y = x^2 + 6x + 10 \quad y = ax^2 + bx + 7/2 \)
\( \frac{dy}{dx} \Big|_{(-2, 2)} = 2x + 6 \Big|_{(-2, 2)} = 2 \); \( \frac{dy}{dx} \Big|_{(1, 2)} = 2ax + b \Big|_{(1, 2)} \)
\( y - 2 = -2(x + 2) \)
\( y - 2 = -2x + 4 \)
\( y + 2x = 6 \)
\( \frac{dy}{dx} = 2a + b \)
\( 2a + b = -2 \quad .......(1) \)
\( (1, 2) \) will lie on the curve
\( 2 = a + b + 7/2 \)
\( a + b = -\frac{3}{2} \quad .......(2) \)
from (1) & (2)
\( a = 1, b = -5/2 \)

Question. The co-ordinates of the point(s) on the graph of the function, \( f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4 \) where the tangent drawn cut off intercepts from the co-ordinate axes which are equal in magnitude but opposite in sign is
(a) \( (2, 8/3) \)
(b) \( (3, 7/2) \)
(c) \( (1, 5/6) \)
(d) None of the options
Answer: (a) \( (2, 8/3) \), (b) \( (3, 7/2) \)
Solution:
\( f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4 \)
\( f'(x) = x^2 - 5x + 7 \mid_{P} \quad P(x_1, y_1) \)
\( f'(x) = x_1^2 - 5x_1 + 7 \)
cuts equal intercepts that means slope = 1
\( x_1^2 - 5x_1 + 7 = 1 \)
\( x_1^2 - 5x_1 + 6 = 0 \)
\( x_1 = 2 , \quad x_1 = 3 \)
\( y_1 = 8/3 , \quad y_1 = 7/2 \)
Point \( (2, 8/3) \), \( (3, 7/2) \)

Question. The co-ordinates of a point on the parabola \( 2y = x^2 \) which is nearest to the point \( (0, 3) \) is
(a) \( (2, 2) \)
(b) \( (-\sqrt{2}, 1) \)
(c) \( (\sqrt{2}, 1) \)
(d) \( (-2, 2) \)
Answer: (a) \( (2, 2) \), (d) \( (-2, 2) \)
Solution:
\( 2y = x^2 \quad P(x_1, y_1) \)
\( \frac{dy}{dx} = \frac{2x}{2} = x \Big|_P = x_1 \)
Equation of normal
\( y - y_1 = -\frac{1}{x_1} (x - x_1) \)
If will pass through \( (0, 3) \)
\( 3 - y_1 = -\frac{1}{x_1} (0 - x_1) \)
\( 3 - y_1 = 1 \implies x_1^2 = 4 \)
\( y_1 = 2 \implies x_1 = 2, -2 \)
point \( (2, 2) \), \( (-2, 2) \)

Question. Consider the curve \( f(x) = x^{1/3} \), then
(a) the equation of tangent at \( (0, 0) \) is \( x = 0 \)
(b) the equation of normal at \( (0, 0) \) is \( y = 0 \)
(c) normal to the curve does not exist at \( (0, 0) \)
(d) \( f(x) \) and its inverse meet at exactly 3 points.
Answer: (a) the equation of tangent at \( (0, 0) \) is \( x = 0 \), (b) the equation of normal at \( (0, 0) \) is \( y = 0 \), (d) \( f(x) \) and its inverse meet at exactly 3 points.
Solution:
Tangent is y-axis at \( (0, 0) \)
\( \implies x = 0 \)
Normal \( y = 0 \)
Exactly three points

Question. The equation of tangents to the curve \( y = \cos (x + y) \), \( -2\pi \le x \le 2\pi \), that are parallel to the line \( x + 2y = 0 \) is/are
(a) \( x + 2y = \pi/2 \)
(b) \( x + 2y = -3\pi/2 \)
(c) \( x - 2y = \pi/2 \)
(d) \( x - 2y = -3\pi/2 \)
Answer: (a) \( x + 2y = \pi/2 \), (b) \( x + 2y = -3\pi/2 \)
Solution:
\( y = \cos (x + y) \)
\( y' = -\sin (x + y) \cdot (1 + y') \)
\( y' = -\frac{\sin (x + y)}{1 + \sin (x + y)} = -\frac{1}{2} \)
\( \sin (x + y) = 1 \)
\( \cos (x + y) = 0 \)
\( y_1 = \cos (x_1 + y_1) = 0 \)
\( \sin (x_1 + y_1) = 1 \implies \sin x_1 = 1 \)
\( x_1 = \frac{\pi}{2}, -\frac{3\pi}{2} \)
Point \( \left( \frac{\pi}{2}, 0 \right) \) and \( \left( -\frac{3\pi}{2}, 0 \right) \)
This two points satisfies.

Question. The normal to the curve \( x = a(\cos \theta + \theta \sin \theta) \), \( y = a(\sin \theta - \theta \cos \theta) \) at any point '\( \theta \)' is such that
(a) It is at a constant distance from the origin
(b) It passes through \( (a\pi/2, -a) \)
(c) It makes angle \( \pi/2 + \theta \) with the x-axis
(d) It passes through the origin
Answer: (a) It is at a constant distance from the origin, (c) It makes angle \( \pi/2 + \theta \) with the x-axis
Solution:
\( x = a(\cos \theta + \theta \sin \theta) \quad y = a(\sin \theta - \theta \cos \theta) \)
\( \frac{dx}{d\theta} = a(-\sin \theta + \sin \theta + \theta \cos \theta) \)
\( \frac{dy}{d\theta} = a(\cos \theta - \cos \theta + \theta \sin \theta) \)
\( \frac{dy}{dx} = \frac{a\theta \sin \theta}{a\theta \cos \theta} = \tan \theta \)
Normal
\( y - a(\sin \theta - \theta \cos \theta) = -\frac{1}{\tan \theta} (x - a(\cos \theta + \theta \sin \theta)) \quad .....(1) \)
\( y - a(\sin \theta - \theta \cos \theta) = -\cot \theta (x - a(\cos \theta + \theta \sin \theta)) \)
\( = \tan \left( \frac{\pi}{2} + \theta \right) (x - a(\cos \theta + \theta \sin \theta)) \)
make are angle of \( \left( \frac{\pi}{2} + \theta \right) \)
Distance from origin of normal (1)
\( d = \left| \frac{-a(\sin \theta - \theta \cos \theta) - a(\cos \theta + \theta \sin \theta)\frac{\cos \theta}{\sin \theta}}{\sqrt{1 + 1/\tan^2 \theta}} \right| \)
\( = a \) which is constant

Question. In the curve \( x = t^2 + 3t - 8 \), \( y = 2t^2 - 2t - 5 \), at point \( (2, -1) \)
(a) length of subtangent is \( 7/6 \)
(b) slope of tangent is \( 6/7 \)
(c) length of tangent is \( \sqrt{85} / 6 \)
(d) None of the options
Answer: (a) length of subtangent is \( 7/6 \), (b) slope of tangent is \( 6/7 \), (c) length of tangent is \( \sqrt{85} / 6 \)
Solution:
\( x = t^2 + 3t - 8 \quad y = 2t^2 - 2t - 5 \)
at point \( (2, -1) \)
\( 2 = t^2 + 3t - 8 \quad -1 = 2t^2 - 2t - 5 \)
\( t = -5, 2 \quad 2t^2 - 2t - 4 = 0 \)
\( t = 2 \quad t = 2 \quad t = -1 \)
\( x_1 = 4 + 6 - 8 \quad y_1 = 8 - 4 - 5 \)
\( x_1 = 2 \quad y_1 = -1 \)
\( \frac{dx}{dt} = 2t + 3 \quad \frac{dy}{dt} = 4t - 2 \)
\( \frac{dy}{dx} = \frac{4t - 2}{2t + 3} \Big|_{t = 2} = \frac{6}{7} \)
\( m = 6/7 \)
\( L_T = \left| \frac{y_1 \sqrt{1 + m^2}}{m} \right| = \left| \frac{-1 \sqrt{1 + \frac{36}{49}}}{6/7} \right| = \frac{\sqrt{85}}{6} \)
Slope of tangent = \( 6/7 \)
\( L_{ST} = \left| \frac{y_1}{m} \right| = \frac{7}{6} \)

Question. If the line, \( ax + by + c = 0 \) is a normal to the curve \( xy = 2 \), then
(a) \( a < 0, b > 0 \)
(b) \( a > 0, b < 0 \)
(c) \( a > 0, b > 0 \)
(d) \( a < 0, b < 0 \)
Answer: (a) \( a < 0, b > 0 \), (b) \( a > 0, b < 0 \)
Solution:
\( by = -ax - c \quad xy = 2 \)
\( y = -\frac{a}{b}x - c \quad y + xy' = 0 \)
\( y' = -\frac{y}{x} \)
Slope of normal = \( \frac{x}{y} \)
\( -\frac{a}{b} = \frac{x_1}{y_1} = \frac{x_1}{2/x_1} = \frac{x_1^2}{2} > 0 \)
LHS always positive.
so RHS should be positive so a, b should have opposite sign.

Question. If the curves \( \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \) & \( y^3 = 16x \) intersect at right angles, then values of a is/are
(a) \( \frac{2}{\sqrt{3}} \)
(b) 2
(c) \( -\frac{2}{\sqrt{3}} \)
(d) not possible
Answer: (a) \( \frac{2}{\sqrt{3}} \), (c) \( -\frac{2}{\sqrt{3}} \)
Solution:
\( \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \); \( y^3 = 16x \)
\( \frac{2x}{a^2} + \frac{2yy'}{4} = 0 \quad 3y^2y' = 16 \)
\( y' = \left. \frac{-4x}{a^2 y} \right|_p \quad y' = \left. \frac{16}{3y^2} \right|_p \)
\( m_1 = -\frac{4x_1}{a^2 y_1} \quad m_2 = y' = \frac{16}{3y_1^2} \)
\( m_1 \times m_2 = -1 \)
\( -\frac{4x_1}{a^2 y_1} \times \frac{16}{3y_1^2} = -1 \)
\( \frac{4}{3a^2} = 1 \)
\( a^2 = \frac{4}{3} \)
\( a = \pm \frac{2}{\sqrt{3}} \)

Question. The equation of normal to the curve \( \left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 (n \in N) \) at the point with abscissa equal to 'a' can be
(a) \( ax + by = a^2 - b^2 \)
(b) \( ax + by = a^2 + b^2 \)
(c) \( ax - by = a^2 - b^2 \)
(d) \( bx - ay = a^2 - b^2 \)
Answer: (a) \( ax + by = a^2 - b^2 \), (c) \( ax - by = a^2 - b^2 \)
Solution:
\( \left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 \quad (n \in N) \)
\( x = a \implies 1 + \left(\frac{y}{b}\right)^n = 2 \)
\( \left(\frac{y}{b}\right)^n = 1 \implies y^n = b^n \)
If n is even: \( y = \pm b \)
If n is odd: \( y = b \)
\( n \to \) even (a, b) & (a, -b)
\( n \to \) odd (a, b)
\( \left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2 \)
\( n\left(\frac{x}{a}\right)^{n-1} \left(\frac{1}{a}\right) + n\left(\frac{y}{b}\right)^{n-1} \left(\frac{1}{b}\right) \frac{dy}{dx} = 0 \)
\( \frac{dy}{dx} = -\frac{b^n}{a^n} \frac{x^{n-1}}{y^{n-1}} \)
\( n \to \) even
\( (a, b) \implies \frac{dy}{dx} = -\left(\frac{b}{a}\right)^n \left(\frac{a}{b}\right)^{n-1} = -\frac{b}{a} = m_T \)
\( M_N = \frac{a}{b} \)
\( y - b = \frac{a}{b} (x - a) \)
\( by - b^2 = ax - a^2 \)
\( ax - by = a^2 - b^2 \)
\( (a, -b) \implies \frac{dy}{dx} = -\left(\frac{b}{a}\right)^n \left(-\frac{a}{b}\right)^{n-1} = \frac{b}{a} = m_T \)
\( M_N = -a/b \)
\( y + b = -\frac{a}{b} (x - a) \)
\( ax + by = a^2 - b^2 \)

Question. Let the parabolas \( y = x^2 + ax + b \) and \( y = x(c - x) \) touch each other at the point \( (1, 0) \). Then
(a) \( a = -3 \)
(b) \( b = 1 \)
(c) \( c = 2 \)
(d) \( b + c = 3 \)
Answer: (a) \( a = -3 \), (d) \( b + c = 3 \)
Solution:
\( y = x^2 + ax + b \quad (1, 0) \text{ satisfies} \)
\( y = x(c - x) \quad 1 + a + b = 0 \)
\( = cx - x^2 \quad c - 1 = 0 \implies c = 1 \)
\( m_1 = \frac{dy}{dx} = 2x + a \Big|_{(1, 0)} = 2 + a \)
\( m_2 = \frac{dy}{dx} = c - 2x \Big|_{(1, 0)} = c - 2 \)
\( 2 + a = c - 2 \)
\( c = 1 \quad a = -3 \)
\( a + b = -1 \)
\( b = -1 - a \)
\( b = -1 + 3 = 2 \)
\( b + c = 3 \)

Question. For the curve represented parametrically by the equation, \( x = 2 \ln \cot t + 1 \) and \( y = \tan t + \cot t \)
(a) tangent at \( t = \pi/4 \) is parallel to x-axis
(b) normal at \( t = \pi/4 \) is parallel to y-axis
(c) tangent at \( t = \pi/4 \) is parallel to the line \( y = x \)
(d) tangent and normal intersect at the point \( (2, 1) \)
Answer: (a) tangent at \( t = \pi/4 \) is parallel to x-axis, (b) normal at \( t = \pi/4 \) is parallel to y-axis
Solution:
\( x = 2 \ln \cot t + 1 \quad y = \tan t + \cot t \)
\( \frac{dx}{dt} = \frac{2}{\cot t} (-\text{cosec}^2 t) \quad \frac{dy}{dt} = \sec^2 t - \text{cosec}^2 t \)
at \( t = \frac{\pi}{4} \)
\( \frac{dx}{dt} = -\frac{2}{\sin^2 t} \tan t \quad \frac{dy}{dt} = 2 - 2 = 0 \)
\( = \frac{-2}{(1/2)} = -4 \)
\( \frac{dy}{dx} = 0 \)
Tangent is parallel to x-axis
Normal is parallel to y-axis

Question. The angle at which the curve \( y = ke^{kx} \) intersects the y-axis is :
(a) \( \tan^{-1}(k^2) \)
(b) \( \cot^{-1}(k^2) \)
(c) \( \sin^{-1}\left(\frac{1}{\sqrt{1+k^4}}\right) \)
(d) \( \sec^{-1}\left(\sqrt{1+k^4}\right) \)
Answer: (b) \( \cot^{-1}(k^2) \), (c) \( \sin^{-1}\left(\frac{1}{\sqrt{1+k^4}}\right) \)
Solution:
\( y = ke^{kx} \) at y-axis
\( m = \frac{dy}{dx} = k^2 e^{kx} \quad x = 0, y = k \)
\( m = k^2 \)
\( \tan \theta = \frac{1}{k^2} \)
\( \theta = \tan^{-1}\left(\frac{1}{k^2}\right) = \cot^{-1} k^2 = \sin^{-1}\left(\frac{1}{\sqrt{1+k^4}}\right) \)

Question. Which of the following pair(s) of curves is/are orthogonal
(a) \( y^2 = 4ax; y = e^{-x/2a} \)
(b) \( y^2 = 4ax; x^2 = 4ay \)
(c) \( xy = a^2; x^2 - y^2 = b^2 \)
(d) \( y = ax; x^2 + y^2 = c^2 \)
Answer: (a) \( y^2 = 4ax; y = e^{-x/2a} \), (c) \( xy = a^2; x^2 - y^2 = b^2 \), (d) \( y = ax; x^2 + y^2 = c^2 \)
Solution:
(A) \( y^2 = 4ax \quad y = e^{-x/2a} \)
\( m_1 = \frac{4a}{2y} = \frac{2a}{y_1} \); \( m_2 = -\frac{1}{2a} e^{-x/2a} = -\frac{1}{2a} y_1 \)
\( m_1 \times m_2 = -1 \)
(B) \( y^2 = 4ax \quad x^2 = 4ay \)
\( y' = \frac{4a}{2y} \quad 2x = 4ay' \implies y' = \frac{2x}{4a} \)
(C) \( xy = a^2 \quad x^2 - y^2 = b^2 \)
\( y + xy' = 0 \quad 2x - 2yy' = 0 \)
\( y' = -\frac{y}{x} \quad y' = \frac{x}{y} \)
\( m_1 \times m_2 = -1 \)
(D) \( y = ax \quad x^2 + y^2 = c^2 \)
\( y' = a \quad 2x + 2yy' = 0 \)
\( m_1 \times m_2 = -\frac{ax}{y} = -1 \quad y' = -\frac{x}{y} \)

Question. If \( y = f(x) \) be the equation of a parabola which is touched by the line \( y = x \) at the point where \( x = 1 \). Then
(a) \( f'(1) = 1 \)
(b) \( f'(0) = f'(1) \)
(c) \( 2f(0) = 1 - f'(0) \)
(d) \( f(0) + f'(0) + f''(0) = 1 \)
Answer: (a) \( f'(1) = 1 \), (c) \( 2f(0) = 1 - f'(0) \)
Solution:
\( y = f(x) \)
Let \( f(x) = ax^2 + bx + c \)
this parabola touches \( y = x \) line at \( (1, 1) \)
that means slope at \( (1, 1) = 1 \)
\( f'(x) = 2ax + b \)
\( 2a + b = 1 \quad ........(1) \)
\( (1, 1) \) will also satisfy the curve
\( 1 = a + b + c \quad ........(2) \)
\( f'(1) = 1 \)
\( 2f(0) = 1 - f'(0) \)
\( f(0) = c \)
LHS = \( 2c \)
RHS = \( 1 - f'(0) \) \quad from (1) & (2) \( a = c \)
\( = 1 - b \)
\( = 2a = 2c \)