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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Easy

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Practice format and chapter quiz

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 1-30 of 164 questions
1 Easy
1 of 164
Analyzing +4 | -1
The complete set of value of 'aa' for which there exists at least one line that is tangent to the graph of the curve y=x3−axy=x^3-a x at one point and normal to the graph at another point is given by
View answer and explanation
Correct option: C
[4/3,∞)[4 / 3, \infty)
Option C ([4/3,∞)[4 / 3, \infty)) is correct because it represents the complete set of values of 'a' for which there exists at least one line that is tangent to the graph of the curve at one point and normal at another.
2 Easy
2 of 164
Analyzing +4 | -1
The point of intersection of the tangents drawn to the curve x2y=1−yx^2 y=1-y at the points where it is met by the curve xy=1−yx y=1-y, is given by
View answer and explanation
Correct option: C
(0,1)(0,1)
Option C ((0,1)) is correct because it is the point of intersection of the tangents to the curve at the points where it intersects another curve.
3 Easy
3 of 164
Application +4 | -1
The curve y−exy+x=0y-e^{x y}+x=0 has a vertical tangent at the point:
View answer and explanation
Correct option: D
(1,0)(1,0)
Option D ((1,0)) is correct because at this point the derivative dx/dy equals zero, indicating a vertical tangent.
4 Easy
4 of 164
Understanding +4 | -1
If normal to the curve y=f(x)y = f\left( x \right) is parallel to xx-axis, then
View answer and explanation
Correct option: C
dxdy=0\frac{{dx}}{{dy}} = 0
Option C (dxdy=0\frac{{dx}}{{dy}} = 0) is correct because if the normal to the curve is parallel to the x-axis, then the tangent to the curve is parallel to the y-axis, implying that dx/dy = 0.
5 Easy
5 of 164
Application +4 | -1
The tangents to the curve x=a(θ−sin⁡θ),  y=a(1+cos⁡θ)x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right) at the points θ=(2k+1)π,  k∈Z\theta = \left( {2k + 1} \right)\pi ,\;k \in Z are parallel to:
View answer and explanation
Correct option: C
y=0y = 0
Option C (y=0y = 0) is correct because the slope dydx\frac{{dy}}{{dx}} is 0 at θ=(2k+1)π\theta = (2k+1)\pi, indicating a horizontal tangent.
6 Easy
6 of 164
Analyzing +4 | -1
The angle of intersection between the curves x2=4(y+1){x^2} = 4\left( {y + 1} \right) and x2=−4(y+1){x^2} = - 4\left( {y + 1} \right) is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the slopes of both curves at the point of intersection are 0, indicating a tangential intersection.
7 Easy
7 of 164
Application +4 | -1
The tangent to the curve y=2x2−x+1y = 2{x^2} - x + 1 at a point PP is parallel to y=3x+4y = 3x + 4, then the coordinates of Pare
View answer and explanation
Correct option: B
(1, 2)
Option B (1, 2) is correct because it is the point where the slope of the tangent to the curve equals the slope of the given line.
8 Easy
8 of 164
Application +4 | -1
The equation of the tangent to curve y=be−x/αy=b e^{-x / \alpha} at the point where it crosses yy-axis is
View answer and explanation
Correct option: D
xa+yb=1\frac{x}{a}+\frac{y}{b}=1
Option D (xa+yb=1\frac{x}{a} + \frac{y}{b} = 1) is correct because it represents the equation of the tangent to the curve y=be−x/ay = be^{-x/a} at the point where it crosses the y-axis (0, b).
9 Easy
9 of 164
Application +4 | -1
The abscissa of the points, where the tangent to 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=−1  and  3x = - 1\;and\;3
Option D (x=−1  and  3x = - 1\;and\;3) is correct because these are the x-coordinates where the derivative of the curve is zero, indicating a horizontal tangent.
10 Easy
10 of 164
Analyzing +4 | -1
Let θ\theta be the acute angle between the tangents to the ellipse x29+y21=1\frac{{{x^2}}}{9} + \frac{{{y^2}}}{1} = 1 and the circle x2+y2=3{x^2} + {y^2} = 3 at their point of intersection in the first quadrant. Then, tan⁡  θ\tan \,\,\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 because calculating the slopes of the tangents and using the angle between two lines formula gives tan⁡θ=23\tan \theta = \frac{2}{{\sqrt 3 }}
11 Easy
11 of 164
Application +4 | -1
The equation of the tangent to the curve y=x+4x2y = x + \frac{4}{{{x^2}}} that is parallel to the x-axis, is
View answer and explanation
Correct option: D
Y=3Y = 3
Option D (Y=3Y = 3) is correct because it represents the equation of the tangent line that is parallel to the x-axis, found by setting the derivative to zero and solving for y.
12 Easy
12 of 164
Application +4 | -1
If f(x)=log⁡x(ln⁡x)f(x)=\log _x(\ln x), then at x=e,f′(x)x=e, f^{\prime}(x) equals-
View answer and explanation
Correct option: D
1/e1 / e
Option D (1/e) is correct because after applying the change of base formula and differentiating, the value of f'(e) is 1/e.
13 Easy
13 of 164
Application +4 | -1
The angle of intersection of the curves y=x2,  6y=7−x3y = {x^2},\;6y = 7 - {x^3} at (1, 1) is
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi }{2}) is correct because the product of the slopes of the tangents at the point of intersection is -1, indicating that the curves intersect at right angles.
14 Easy
14 of 164
Analyzing +4 | -1
If y=cos⁡−1(cos⁡x)y=\cos ^{-1}(\cos x), then y′(x)y^{\prime}(x) is equal to
View answer and explanation
Correct option: A
1 for all xx
Option A (1 for all xx) is correct because the derivative of the inverse cosine of cosine x simplifies to 1.
15 Easy
15 of 164
Application +4 | -1
The equation of the normal line to the curve y=xlog⁡exy = x{\log _e}x parallel to 2x−2y+3=02x - 2y + 3 = 0 is
View answer and explanation
Correct option: C
x−y=3e−2x - y = 3{e^{ - 2}}
Option C (x−y=3e−2x - y = 3{e^{ - 2}}) is correct because it represents the normal line to the curve that is parallel to the given line, found by equating slopes and solving for the point of tangency.
16 Easy
16 of 164
Analyzing +4 | -1
A curve with equation of the form y=ax4+bx3+c+cx+dy=a x^4+b x^3+c+c x+d has zero gradient at the point (0,1)(0,1) and also touches the xx-axis at the point (−1,0)(-1,0) then the values of the xx for which the curve has a negative gradient are:
View answer and explanation
Correct option: C
x<−1x < -1
Option C (x<−1x < -1) is correct because it identifies the range of x-values where the derivative of the curve is negative, indicating a negative gradient.
17 Easy
17 of 164
Understanding +4 | -1
The normal to a given curve is parallel to xx-axis, if
View answer and explanation
Correct option: C
dxdy=0\frac{d x}{d y}=0
Option C (dxdy=0\frac{d x}{d y}=0) is correct because if the normal is parallel to the x-axis, the tangent is perpendicular to the x-axis, implying dxdy=0\frac{dx}{dy} = 0.
18 Easy
18 of 164
Application +4 | -1
The slope of tangent to the curve x=t2+3t−8x=t^2+3 t-8, y=2t2−2t−5y=2 t^2-2 t-5 at the point (2,−1)(2,-1) is
View answer and explanation
Correct option: B
67\frac{6}{7}
Option B (67\frac{6}{7}) is correct because it is the value of dydx\frac{dy}{dx} at t=2, which is found by solving for t and then substituting into the derivative.
19 Easy
19 of 164
Application +4 | -1
At any point of a curve  subnormal  subtangent \sqrt{\frac{\text { subnormal }}{\text { subtangent }}} is equal to
View answer and explanation
Correct option: C
Slope of the tangent at that point
Option C (Slope of the tangent at that point) is correct because the square root of the ratio of subnormal to subtangent is equal to the slope of the tangent.
20 Easy
20 of 164
Application +4 | -1
Let CC be a curve given by y=1+4x−3y=1+\sqrt{4 x-3}, x>34x > \frac{3}{4}. If PP is a point on C\mathrm{C} such that the tangent at PP has slope 23\frac{2}{3}, then a point through which the normal at PP passes, is
View answer and explanation
Correct option: B
(1,7)(1,7)
Option B (1,7) is correct because the normal at P(3,4) passes through the point (1,7).
21 Easy
21 of 164
Application +4 | -1
The equation of one of the tangents to the curve y=cos⁡(x+y),−2π≤x≤2πy=\cos (x+y),-2 \pi \leq x \leq 2 \pi that is parallel to the line x+2y=0x+2 y=0, is
View answer and explanation
Correct option: B
x+2y=π/2x+2 y=\pi / 2
The correct answer is x+2y=π2x+2y = \frac{\pi}{2} because it represents the tangent to the curve y=cos⁡(x+y)y = \cos(x+y) that is parallel to the line x+2y=0x+2y=0.
22 Easy
22 of 164
Application +4 | -1
The distance between the origin and the tangent to the curve y=e2x+x2y=e^{2 x}+x^2 drawn at the point x=0x=0 is
View answer and explanation
Correct option: A
15\frac{1}{\sqrt{5}}
Option A (15\frac{1}{\sqrt{5}}) is correct because it represents the perpendicular distance from the origin to the tangent line calculated at x=0.
23 Easy
23 of 164
Application +4 | -1
he length of the subnormal to the parabola y2=4  ax{y^2} = 4\;ax at any point is equal to
View answer and explanation
Correct option: D
2a2a
Option D (2a2a) is correct because the length of the subnormal to the parabola y2=4ax{y^2} = 4ax is constant and equal to 2a2a.
24 Easy
24 of 164
Application +4 | -1
If the curve y=ax2+bx+c,y = a{x^2} + bx + c, x∈R,x \in R, passes through the point (1,  2)(1,\,\,2) and the tangent line to this curve at origin is y=x,y = x, then the possible values of a,  b,  ca,\,\,b,\,\,c are
View answer and explanation
Correct option: C
a=1,    b=1,   c=0a = 1,\,\,\,\,b = 1,\,\,\,c = 0
Option C (a=1,   b=1,   c=0a = 1,\,\,\,b = 1,\,\,\,c = 0) is the correct set of values because the curve passes through (1,2) and has a tangent line y=x at the origin.
25 Easy
25 of 164
Application +4 | -1
The slope of the tangent to the curve y=∫0x11+t3dty=\int\limits_{0}^x \frac{1}{1+t^3} d t at the point where x=1x=1, is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because the slope of the tangent is found by differentiating the given integral and evaluating it at x=1.
26 Easy
26 of 164
Application +4 | -1
The point on the curve y2=x{y^2} = x, the tangent at which makes an angle 450{45^0} with xx-axis is
View answer and explanation
Correct option: A
(14,12)\left( {\frac{1}{4},\frac{1}{2}} \right)
Option A ((14,12)\left( {\frac{1}{4},\frac{1}{2}} \right)) is correct because it represents the point on the curve where the tangent has a slope of 1, corresponding to a 45-degree angle with the x-axis.
27 Easy
27 of 164
Analyzing +4 | -1
The point in the interval [0,  2π]\left[ {0,\;2\pi } \right] where f(x)=exsin⁡xf\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 maximum slope occurs where the second derivative is zero and the third derivative is negative.
28 Easy
28 of 164
Analyzing +4 | -1
If the line ax+by+c=0ax + by + c = 0 is a tangent to the curve xy=4,xy = 4, then
View answer and explanation
Correct option: C
a<0,b<0a < 0,b < 0
Option C (a<0,b<0a < 0, b < 0) is correct because for the line to be tangent to the curve, the product of a and b must be positive, and since the slope is negative, both a and b must be negative.
29 Easy
29 of 164
Application +4 | -1
The equation of the normal to the curve y=e−2∣x∣y=e^{-2|x|} at the point where the curve cuts the line x=−12x=-\frac{1}{2}, is
View answer and explanation
Correct option: A
2e(ex+2y)=4−e22 e(e x+2 y)=4-e^2
Option A (2e(ex+2y)=4−e22 e(e x+2 y)=4-e^2) is correct because it represents the equation of the normal to the curve at the given point, derived using the derivative and the negative reciprocal of the slope.
30 Easy
30 of 164
Application +4 | -1
The equation of the tangent to the curve x2−2xy+y2+2x+y−6=0{x^2} - 2xy + {y^2} + 2x + y - 6 = 0 at (2,2) is
View answer and explanation
Correct option: A
2x+y−6=02x + y - 6 = 0
Option A (2x+y−6=02x + y - 6 = 0) is correct because it represents the tangent line to the curve at the point (2,2) with the calculated slope.
Revision notes and key takeaways

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Key Takeaways

  • ✓ This set contains 164 single-correct-answer MCQs on TANGENTS & NORMALS.
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