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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 181-210 of 281 questions

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181 Medium
181 of 281
Analyzing +4 | -1
The line xa+yb=2\frac{x}{a} + \frac{y}{b} = 2 touches the curve (xa)n+(yb)n=2{\left( {\frac{x}{a}} \right)^n} + {\left( {\frac{y}{b}} \right)^n} = 2 at the point (a,b) for
View answer and explanation
Correct option: C
Any n∈Rn \in R
Option C (Any n∈Rn \in R) is correct because the tangent to the curve at (a, b) is the same regardless of the value of n.
182 Easy
182 of 281
Application +4 | -1
The tangent and the normal drawn to the curve y=x2−x+4y = {x^2} - x + 4 at P(1,  4)P\left( {1,\;4} \right) cut the xx- axis at AA and BB respectively. If the length of the subtangent drawn to the curve at PP is equal to the length of the subnormal, then the area of the triangle PABPAB in sq units is
View answer and explanation
Correct option: D
16
Option D (16) is correct because it represents the area of the triangle formed by the tangent and normal to the curve at the given point, calculated using coordinate geometry.
183 Medium
183 of 281
Analyzing +4 | -1
Angle between y2=x{y^2} = x and x2=y{x^2} = y at the origin is
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi }{2}) is correct because the tangents at the origin are the x and y axes, which are perpendicular.
184 Hard
184 of 281
Analyzing +4 | -1
If the tangent to the conic, y−6=x2y-6=x^2 at (2,10)(2,10) touches the circle, x2+y2+8x−2y=kx^2+y^2+8 x-2 y=k (for some fixed kk ) at a point (α,β)(\alpha, \beta); then (α,β)(\alpha, \beta) is :
View answer and explanation
Correct option: D
(−817,217)\left(-\frac{8}{17}, \frac{2}{17}\right)
Option D ((−817,217)\left(-\frac{8}{17}, \frac{2}{17}\right)) is correct. The tangent to the conic touches the circle at a point that can be found by equating the slopes and solving the system of equations.
185 Easy
185 of 281
Application +4 | -1
The equation of the normal to the parabola, x2=8yx^2=8 y at x=4x=4 is
View answer and explanation
Correct option: A
x+y=6x+y=6
Option A (x+y=6x+y=6) is correct because it represents the equation of the normal line to the parabola at the given point.
186 Easy
186 of 281
Analyzing +4 | -1
The line xa+yb=1\frac{x}{a} + \frac{y}{b} = 1 touches the curve y=be−x/ay = b{e^{ - x/a}} at the point
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the point of tangency is (0, b), which is not represented by any of the other options.
187 Medium
187 of 281
Application +4 | -1
If the straight line joining the points (0,3)(0,3) and (5,−2)(5,-2) is tangent to the curve y=cx+1y=\frac{c}{x+1}; then the value of cc is :
View answer and explanation
Correct option: C
4
Option C (4) is correct because the tangent condition and the equation of the line allow us to solve for 'c', which equals 4.
188 Easy
188 of 281
Analyzing +4 | -1
Let y=f(x)y=f(x) be the equation of a parabola which is touched by the line y=xy=x at the point where x=1x=1. Then,
View answer and explanation
Correct option: D
2f(0)=1−f′(0)2 f(0)=1-f^{\prime}(0)
Option D (2f(0)=1−f′(0)2 f(0)=1-f^{\prime}(0)) is correct because it is derived from the conditions that the parabola is tangent to the line y=x at x=1.
189 Easy
189 of 281
Application +4 | -1
The angle between the tangents to the curve y=x2−5x+6y=x^2-5 x+6 at the point (2,0)(2,0) and (3,0)(3,0), is
View answer and explanation
Correct option: B
π/2\pi / 2
Option B (π/2\pi / 2) is correct because the product of the slopes of the tangents at the given points is -1, indicating perpendicularity.
190 Easy
190 of 281
Analyzing +4 | -1
If y(α)=2(tan⁡α+cot⁡α1+tan⁡2α)+1sin⁡2α,α∈(3π4,π),y(\alpha ) = \sqrt {2\left( {\frac{{\tan \alpha + \cot \alpha }}{{1 + {{\tan }^2}\alpha }}} \right) + \frac{1}{{{{\sin }^2}\alpha }}} ,\alpha \in \left( {\frac{{3\mathrm{\pi }}}{4},\mathrm{\pi }} \right), then dydα\mathrm{ }\frac{{dy}}{{d\alpha }} at α=5π6\alpha = \frac{{5\mathrm{\pi }}}{6} is
View answer and explanation
Correct option: D
4
Option D (4) is correct because after simplifying the expression for y(α\alpha) and differentiating, we find that dydα\frac{{dy}}{{d\alpha}} at α=5π6\alpha = \frac{{5\pi}}{{6}} is equal to 4.
191 Medium
191 of 281
Application +4 | -1
Angle between the tangents to the curve y=x2−5x+6y = {x^2} - 5x + 6 at the point (2, 0) and (3, 0) is
View answer and explanation
Correct option: A
π2\frac{\pi }{2}
Option A (π2\frac{\pi }{2}) is correct because the product of the slopes of the tangents at the given points is -1, indicating perpendicularity.
192 Easy
192 of 281
Application +4 | -1
The length of the normal at point 'tt' of the curve x=a(t+sin⁡t),  y=a(1−cos⁡t)x = a\left( {t + \sin t} \right),\;y = a\left( {1 - \cos t} \right) is
View answer and explanation
Correct option: C
2asin⁡(t/2)tan⁡(t/2)2a\sin \left( {t/2} \right)\tan \left( {t/2} \right)
Option C (2asin⁡(t/2)tan⁡(t/2)2a\sin \left( {t/2} \right)\tan \left( {t/2} \right)) is correct because it represents the length of the normal calculated using the derivative of the parametric equations and trigonometric identities.
193 Easy
193 of 281
Application +4 | -1
The equation of the normal to the curve y4=ax3  {y^4} = a{x^3}\; at (a, a) is
View answer and explanation
Correct option: C
4x+3y=7a4x + 3y = 7a
Option C (4x+3y=7a4x + 3y = 7a) is correct because it represents the equation of the normal line to the curve at the specified point, calculated using the derivative and the negative reciprocal of the tangent's slope.
194 Medium
194 of 281
Application +4 | -1
The equation of a normal to the curve, sin⁡y=xsin⁡(π3+y)\sin \mathrm{y}=\mathrm{x} \sin \left(\frac{\pi}{3}+y\right) at x=0\mathrm{x}=0, is :
View answer and explanation
Correct option: B
2x+3y=02 x+\sqrt{3} y=0
Option B (2x+3y=02 x+\sqrt{3} y=0) is correct because it represents the equation of the normal to the curve at x=0, derived from the derivative and normal slope.
195 Easy
195 of 281
Application +4 | -1
The points on the curve y=12x−x3y=12 x-x^3 at which the gradient is zero are
View answer and explanation
Correct option: D
(2,16),(−2,−16)(2,16),(-2,-16)
Option C ((2,−16),(−2,16)(2,-16),(-2,16)) is correct because setting the derivative of the curve equal to zero and solving for x gives x=±\pm2, and substituting these values back into the original equation yields the corresponding y-values.
196 Easy
196 of 281
Application +4 | -1
Let slope of the tangent line to a curve at any point P (x, y)P\,(x,\,y) be given by xy2+yx\frac{{x{y^2} + y}}{x}. If the curve intersects the line x+2y=4x + 2y = 4 at x=−2x = - 2, then the value of yy, for which the point (3, y)(3,\,y) lies on the curve, is
View answer and explanation
Correct option: C
−1819- \frac{{18}}{{19}}
Option C (−1819- \frac{{18}}{{19}}) is correct because it is the value of y that satisfies the differential equation and the given condition.
197 Easy
197 of 281
Application +4 | -1
The subtangent, ordinate and subnormal to the parabola y2=4ax{y^2} = 4ax at a point (different from the origin) are in
View answer and explanation
Correct option: B
GP
Option B (GP) is correct because the ordinate squared equals the product of the subtangent and subnormal, indicating a geometric progression.
198 Easy
198 of 281
Application +4 | -1
If at each point on the curve y=x3−ax2+x+1y=x^3-a x^2+x+1 the tangent is inclined at an acute angle with the positive direction of xx-axis, then
View answer and explanation
Correct option: C
∣a∣≤3|a| \leq \sqrt{3}
Option C (∣a∣≤3|a| \leq \sqrt{3}) is correct because the condition for the tangent to be inclined at an acute angle with the positive x-axis leads to the discriminant of the quadratic being non-positive, resulting in this inequality.
199 Medium
199 of 281
Analyzing +4 | -1
The normal to the curve represented parametrically by x=a(cos⁡θ+θsin⁡θ) and y=a(sin⁡θ−θcos⁡θ)x = a\left( {\cos \theta + \theta \sin \theta } \right)\,and\,y = a\left( {\sin \theta - \theta \cos \theta } \right) at any point θ\theta, is such that it
View answer and explanation
Correct option: B
Is at a constant distance from the origin
Option B (Is at a constant distance from the origin) is correct because the equation of the normal simplifies to a form where the distance from the origin is constant.
200 Medium
200 of 281
Application +4 | -1
The curve x+y−ℓn(x+y)=2x+5x+y-\ell \mathrm{n}(x+y)=2 x+5 has a vertical tangent at the point (α,β)(\alpha, \beta). Then α+β\alpha+\beta is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because for a vertical tangent, dydx\frac{{dy}}{{dx}} must be undefined, which occurs when the denominator of dydx\frac{{dy}}{{dx}} is zero, leading to α+β=1\alpha + \beta = 1.
201 Easy
201 of 281
Application +4 | -1
If the rate of change of sine of an angle θ\theta is kk, then the rate of change of its tangent is
View answer and explanation
Correct option: B
1k2\frac{1}{{{k^2}}}
Option B (1k2\frac{1}{{{k^2}}}) is correct because it is derived by finding the derivative of tan(θ\theta) with respect to θ\theta and using the given rate of change of sin(θ\theta).
202 Medium
202 of 281
Application +4 | -1
If y=4x−5y = 4x - 5 is tangent to the curve y2=px3+q{y^2} = p{x^3} + q at (2, 3) then (p,q) is
View answer and explanation
Correct option: D
(2,  −7)\left( {2,\; - 7} \right)
Option D (2,  −7)\left( {2,\; - 7} \right) is correct because these values of p and q satisfy both the tangency condition and the point lying on the curve.
203 Easy
203 of 281
Application +4 | -1
The lengths of tangent, subtangent, normal and subnormal for the curve y=x2+x−1y=x^2+x-1 at (1,1)(1,1) are A,B,CA, B, C and DD respectively, then their increasing order is
View answer and explanation
Correct option: D
B,A,D,CB, A, D, C
Option A (B,D,A,CB, D, A, C) is correct because, after calculating the lengths of the tangent, subtangent, normal, and subnormal, their increasing order is indeed B < A < D < C.
204 Medium
204 of 281
Application +4 | -1
The abscissa of the points, where the tangent to the curve y=x3−3x2−9x+5y = {x^3} - 3{x^2} - 9x + 5 is parallel to xx-axis, are
View answer and explanation
Correct option: D
x=−1x = - 1 and 3
Option D (x=−1x = - 1 and 3) is correct because these are the x-coordinates where the derivative of the curve is zero, indicating a horizontal tangent line.
205 Medium
205 of 281
Application +4 | -1
If y=4x−5y = 4x - 5 is a tangent to the curve y2=px3+q{y^2} = p{x^3} + q at (2, 3), then
View answer and explanation
Correct option: A
p=2,  q=−7p = 2,\;q = - 7
Option A (p=2,  q=−7p = 2,\;q = - 7) is correct because these values satisfy both the tangency condition and the point lying on the curve.
206 Easy
206 of 281
Application +4 | -1
The equation of the normal to the curve y=sin⁡xy=\sin x at (0,0)(0,0), is
View answer and explanation
Correct option: C
x+y=0x+y=0
Option C (x+y=0x+y=0) is correct because the slope of the normal at (0,0) is -1, and the line passes through the origin.
207 Medium
207 of 281
Application +4 | -1
The equation of the tangent to the curve y=4xex at (−1,−4e)y = 4x{e^x}\,\mathrm{at}\,\left( { - 1,\frac{{ - 4}}{e}} \right) is
View answer and explanation
Correct option: B
y=−4ey = - \frac{4}{e}
Option C (4) is correct because it satisfies the condition that the sum of the squares of the intercepts made by the tangent to the curve at the given point is 2.
208 Medium
208 of 281
Application +4 | -1
A tangent is drawn at the point MM with abscissa unity on the curve y=(5−x23)3y=\sqrt{\left(5-x^{\frac{2}{3}}\right)^3}. If length of segment of tangent intercepted between the co-ordinate axes is N\sqrt{{N}}, then N{N} is equal to
View answer and explanation
Correct option: D
125
Option D (125) is correct. By finding the derivative, the tangent equation, and then the intercepts, the length of the segment is calculated to be \125\sqrt{{125}}\$.
209 Hard
209 of 281
Application +4 | -1
Tangent is drawn to ellipse x227+y2=1 at (33cos⁡θ,sin⁡θ) (where θ∈(0,  π/2))\frac{{{x^2}}}{{27}} + {y^2} = 1\,at\,\left( {3\sqrt 3 \cos \theta ,\sin \theta } \right)\,(where\,\theta \in \left( {0,\;\pi /2} \right)). Then the value of θ\theta such that sum of intercepts on axes made by this tangent is minimum, is
View answer and explanation
Correct option: B
π6\frac{\pi }{6}
Option B π6\frac{\pi }{6} is correct because it minimizes the sum of the intercepts on the axes, found by setting the derivative of the sum equal to zero.
210 Medium
210 of 281
Application +4 | -1
A curve passes through the point (2,0)(2,0) and the slope of the tangent at any point (x,y)(x, y) is x2−2xx^2-2 x for all value of xx. The point of maximum ordinate on the curve is
View answer and explanation
Correct option: A
(0,4/3)(0,4 / 3)
Option A ((0,4/3)) is correct because the point of maximum ordinate is (0, 4/3), which is found by integrating the slope function and finding the critical points.
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