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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 211-240 of 281 questions

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211 Easy
211 of 281
Application +4 | -1
Equation of normal to x = 2et,y=e−t at t=0x{\text{ = }}2{e^t},y = {e^{ - t}}{\text{ }}at{\text{ }}t = 0 is
View answer and explanation
Correct option: C
2x−y−3=02x - y - 3 = 0
Option C (2x−y−3=02x - y - 3 = 0) is correct because it is the equation of the normal line to the curve at the given point, calculated using derivatives.
212 Easy
212 of 281
Application +4 | -1
If the function f(x)=x2+α/xf(x)=x^2+\alpha / x has a local minimum at x=2x=2, then the value of α\alpha is
View answer and explanation
Correct option: C
16
Option C (16) is correct because the value of alpha is 16, which is found by setting the first derivative of the function equal to zero at x=2 and solving for alpha.
213 Easy
213 of 281
Application +4 | -1
The gradient of the curve passing through (4,0)(4,0) is given by dydx−yx+5x(x+2)(x−3)=0\frac{dy}{dx}-\frac{y}{x}+\frac{5x}{\left( x+2 \right)\left( x-3 \right)}=0, if the point (5,a)(5,a) lies on the curve, then the value of aa is
View answer and explanation
Correct option: C
5 log⁡7125\,\log \frac{7}{12}
The correct answer is D (5 log⁡1275\,\log \frac{12}{7}) because it is derived by solving the given differential equation and applying the provided conditions.
214 Medium
214 of 281
Application +4 | -1
If the sum of the squares of the intercepts on the axes cut off by the tangent to the curve x1/3+y1/3=a1/3{x^{1/3}} + {y^{1/3}} = {a^{1/3}} (with a>0a > 0) at P(a/8,  a/8)P\left( {a/8,\;a/8} \right) is 2, then a=a =
View answer and explanation
Correct option: C
4
Option C (xb\frac{x}{b}) is correct because it represents the acceleration of the particle, derived by differentiating the given equation with respect to time.
215 Medium
215 of 281
Application +4 | -1
If the line y=2x−4k2y=\sqrt{2} x-4 k \sqrt{2} is normal chord of parabola y2=2xy^2=2 x whose length is aa(a∈N)a \sqrt{a}(a \in N), then the value of (a+1k)\left(a+\frac{1}{k}\right) is
View answer and explanation
Correct option: A
5
Option A (5) is correct because by finding the normal chord and its length, we can determine the values of a and k, and then calculate (620/(p+q)) where p/q is the probability that a < b.
216 Easy
216 of 281
Analyzing +4 | -1
The equation of the tangent to the curve y=(2x−1)e2(1−x)y=(2 x-1) e^{2(1-x)} at the point of its maximum, is
View answer and explanation
Correct option: A
y−1=0y-1=0
Option A (y−1=0y-1=0) is correct because it represents the equation of the tangent to the curve at its maximum point.
217 Easy
217 of 281
Analyzing +4 | -1
The angle between the curves y2=4x+4{y^2} = 4x + 4 and y2=36(9−x){y^2} = 36\left( {9 - x} \right) is
View answer and explanation
Correct option: D
900{90^0}
Option D (900{90^0}) is correct because the product of the derivatives of the two curves at their intersection points is -1, indicating perpendicularity.
218 Easy
218 of 281
Application +4 | -1
If the curve y=ax2+bx+cy=a x^2+b x+c passes through the point (1,2)(1,2) such that the slope of the tangent at the origin is equal to 1, then (a,b,c)=(a, b, c)=
View answer and explanation
Correct option: B
(1,1,0)(1,1,0)
Option B ((1,1,0)) is correct because substituting the point (1,2) and the slope of the tangent at the origin into the curve equation yields these values for a, b, and c.
219 Medium
219 of 281
Analyzing +4 | -1
If β\beta is one of the angles between the normals to the ellipse, x2+3y2=9x^2+3 y^2=9 at the points (3cos⁡θ,3sin⁡θ)(3 \cos \theta, \sqrt{3} \sin \theta) and (−3sin⁡θ,3cos⁡θ);∈(0,π2);(-3 \sin \theta, \sqrt{3} \cos \theta) ; \in\left(0, \frac{\pi}{2}\right) ; then 2cot⁡βsin⁡2θ\frac{2 \cot \beta}{\sin 2 \theta} is equal to
View answer and explanation
Correct option: B
23\frac{2}{\sqrt{3}}
Option B (23\frac{2}{\sqrt{3}}) is correct. By finding the slopes of the normals at the given points and using the tangent formula, the expression simplifies to 23\frac{2}{\sqrt{3}}.
220 Easy
220 of 281
Analyzing +4 | -1
The angle between the curves y=axy = {a^x} and y=bxy = {b^x} is equal to
View answer and explanation
Correct option: D
tan⁡−1(∣log⁡a−log⁡b1+log⁡alog⁡b∣  )  {\tan ^{ - 1}}\left( {\left| {\frac{{\log a - \log b}}{{1 + \log a\log b}}} \right|\;} \right)\;
Option D (tan⁡−1(∣log⁡a−log⁡b1+log⁡alog⁡b∣;);{\tan ^{ - 1}}\left( {\left| {\frac{{\log a - \log b}}{{1 + \log a\log b}}} \right|\\;} \right)\\;) is correct because it represents the angle between the curves, calculated using the derivatives at the point of intersection.
221 Easy
221 of 281
Application +4 | -1
Let y be the number of people in a village at time tt. Assume that the rate of change of the population is proportional to the number of people in the village at any time and further assume that the population never increase in time. Then, the population of the village at any fixed tt is given by
View answer and explanation
Correct option: B
y=cekt,y = c{e^{kt}}, for some constants c≥0c \ge 0 and k≤0k \le 0
Option B (y=cekt,y = c{e^{kt}}, for some constants c≥0c \ge 0 and k≤0k \le 0) is correct because it represents exponential decay, which aligns with a population that never increases.
222 Easy
222 of 281
Application +4 | -1
The curves x=y2x=y^2 and xy=a3x y=a^3 cut orthogonally at a point, then a=a=
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (1/2) is correct because the value of 'a' for the curves to cut orthogonally is calculated to be 1/2.
223 Medium
223 of 281
Application +4 | -1
If the tangent at the point (1,1)(1,1) of the curve y2−x(2−x)2y^2-x(2-x)^2 meets the curve again at the point P\mathrm{P}, then P\mathrm{P} is ...... .
View answer and explanation
Correct option: C
(94,38)\left(\frac{9}{4}, \frac{3}{8}\right)
Option C ((9/4, 3/8)) is correct because it is the point where the tangent at (1,1) intersects the curve again, found by solving the system of equations.
224 Hard
224 of 281
Application +4 | -1
The area of a triangle formed by a tangent to the curve 2xy=a22 x y=a^2 and the coordinate axes, is
View answer and explanation
Correct option: B
a2a^2
Option B (a2a^2) is correct because the area of the triangle formed by the tangent and coordinate axes is calculated to be a2a^2.
225 Easy
225 of 281
Application +4 | -1
The curve y=ax3+bx2+cx+5y=a x^3+b x^2+c x+5 touches the x\mathrm{x}-axis at P(−2,0)\mathrm{P}(-2,0) and cut the y\mathrm{y}-axis at a point Q\mathrm{Q} where its gradient is 3 . Then, a+b+c=a+b+c=
View answer and explanation
Correct option: B
7/47 / 4
Option B (7/47 / 4) is correct because it is obtained by using the given conditions to find the coefficients of the cubic polynomial and then calculating a+b+ca+b+c.
226 Medium
226 of 281
Application +4 | -1
If the curve y=2x3+ax2+bx+cy = 2{x^3} + a{x^2} + bx + c passes through the origin and the tangents drawn to it at x=−1x = - 1 and x=2x = 2 are parallel to the xx-axis, then the values of a,ba,b and care respectively
View answer and explanation
Correct option: B
−3,−12and0- 3, - 12\mathrm{ }and\mathrm{ }0
Option B ( - 3, - 12 and 0) is correct because these values satisfy the given conditions: the curve passes through the origin, and the tangents at x = -1 and x = 2 are parallel to the x-axis.
227 Medium
227 of 281
Application +4 | -1
The point on the curve y=2x2−6x−4y = 2{x^2} - 6x - 4 at which the tangent is parallel to the xx-axis, is
View answer and explanation
Correct option: D
(32,  −172)\left( {\frac{3}{2},\; - \frac{{17}}{2}} \right)
Option D ((32\frac{3}{2},\\; - 172\frac{{17}}{2})) is correct because it represents the point on the curve where the derivative (slope of the tangent) is zero, indicating a horizontal tangent.
228 Easy
228 of 281
Application +4 | -1
The equation of the tangent to the curve x=tcos⁡t,  y=tsin⁡tx = t\cos t,\;y = t\sin t at the origin is
View answer and explanation
Correct option: B
y=0y = 0
Option B (y=0y = 0) is correct because at the origin (t=0), the slope dydx\frac{{dy}}{{dx}} is 0, indicating a horizontal tangent.
229 Hard
229 of 281
Application +4 | -1
The length of normal to the curve x=a(θ+sin⁡θ)x=a(\theta+\sin \theta), y=a(1−cos⁡θ)y=a(1-\cos \theta) at the point θ=π/2\theta=\pi / 2 is
View answer and explanation
Correct option: C
2a\sqrt{2} a
Option C (2a\sqrt{2} a) is the correct answer because it is the calculated length of the normal to the curve at the specified point.
230 Medium
230 of 281
Analyzing +4 | -1
The normal to the curve x=a(cos⁡θ+θsin⁡θ),  y=a(sin⁡θ−θcos⁡θ)x = a\left( {\cos \theta + \theta \sin \theta } \right),\;y = a\left( {\sin \theta - \theta \cos \theta } \right) at any point θ\theta is such that
View answer and explanation
Correct option: A
It is at a constant distance from the origin
Option A (It is at a constant distance from the origin) is correct because the equation of the normal can be simplified to a form where the distance from the origin is constant.
231 Easy
231 of 281
Application +4 | -1
Point of contact of tangents to the curve y2−2x3−4y+8=0y^{2}-2 x^{3}-4 y+8=0 from the point (1,2)(1,2) is/are :
View answer and explanation
Correct option: A
(2,2±3)(2,2 \pm \sqrt{3})
Option A is correct because the points of contact are found by solving the tangent equation and the curve equation simultaneously.
232 Medium
232 of 281
Application +4 | -1
If the curves x2a2+y212=1  \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{12}} = 1\; and y3=8x  {y^3} = 8x\; intersect at right angle then the value of a2{a^2} is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because the value of a2{a^2} is 4 when the given curves intersect at right angles, which is found by equating the product of their slopes to -1.
233 Medium
233 of 281
Application +4 | -1
Which one of the following curves is the orthogonal trajectory of straight lines passing through a fixed point (a,b)(a, b) ?
View answer and explanation
Correct option: D
(x−a)2+(y−b)2=k(x-a)^{2}+(y-b)^{2}=k
Option D ((x−a)2+(y−b)2=k(x-a)^{2}+(y-b)^{2}=k) is correct because straight lines passing through a fixed point (a, b) are normals to a circle centered at (a, b), making the circle the orthogonal trajectory.
234 Medium
234 of 281
Application +4 | -1
The point of inflexion for the curve y=x5/2y = {x^{5/2}} is
View answer and explanation
Correct option: B
(0,0)
Option B ((0,0)) is correct because it's the point where the second derivative changes sign, indicating a point of inflection.
235 Easy
235 of 281
Application +4 | -1
The number of points on the curve y=x3−2x2+x−2y = {x^3} - 2{x^2} + x - 2 where tangents are parallel to xx-axis, is
View answer and explanation
Correct option: C
2
The correct answer is 2 because the derivative of the given curve, when set to zero (indicating a tangent parallel to the x-axis), yields two distinct real roots.
236 Easy
236 of 281
Analyzing +4 | -1
The curve y=x15y=x^{\frac{1}{5}} at (0,0)(0,0) has
View answer and explanation
Correct option: B
a horizontal tangent (parallel to xx aria)
Option A (a vertical tangent (parallel to y-axis)) is correct because the derivative approaches infinity at (0,0), indicating a vertical tangent.
237 Medium
237 of 281
Analyzing +4 | -1
The point in the interval   [0  ,2π]\;\left[ {0\;,2\pi } \right], where f(x)=ex  sin⁡x  f\left( x \right) = {e^x}\;\sin x\; has maximum slope, is
View answer and explanation
Correct option: B
π2\frac{\pi }{2}
Option B (π2\frac{\pi}{2}) is correct because the second derivative of the function is zero at this point, and the third derivative is negative, indicating a maximum slope.
238 Hard
238 of 281
Application +4 | -1
The equation of the one of the tangents to the curve y=cos⁡(x+y),  −2π≤x≤2πy = \cos \left( {x + y} \right),\; - 2\pi \le x \le 2\pi that is parallel to the line x+2y=0x + 2y = 0, is
View answer and explanation
Correct option: B
x+2  y=π2x + 2\;y = \frac{\pi }{2}
Option D (B,  A,  D,  CB,\;A,\;D,\;C) is correct because it represents the increasing order of the lengths of subtangent, tangent, subnormal, and normal.
239 Medium
239 of 281
Application +4 | -1
The intercepts made by the tangent to the curve y=∫0x∣t∣dty = \int \limits_0^x \left| t \right|dt, which is parallel to the line y=2xy = 2x, on yy-axis are equal to
View answer and explanation
Correct option: B
−2,2- 2,2
Option B ( - 2,2) is correct because these are the y-intercepts of the tangent lines to the curve that are parallel to the line y = 2x, found by equating the derivatives and solving for the points of tangency.
240 Easy
240 of 281
Application +4 | -1
The subtangent at any point on the curve xmyn=am+nx^m y^n=a^{m+n} varies as
View answer and explanation
Correct option: C
Abscissa
Option C (Abscissa) is correct because the subtangent varies directly with the abscissa (x-coordinate).
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