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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 31-60 of 281 questions
31 Medium
31 of 281
Analyzing +4 | -1
The equation of the tangent to curve y(2x−1)e2(1−x)y\left( {2x - 1} \right){e^{2\left( {1 - x} \right)}} at the points 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 the tangent to the curve at its maximum point (1,1) is a horizontal line, hence y=1.
32 Medium
32 of 281
Application +4 | -1
Let f(x) and g(x)f\left( x \right)\,and\,g\left( x \right) be differentiable for 0≤x≤1,0 \le x \le 1,, such that f(0)=2,g(0)=0,f(1)=6.f\left( 0 \right) = 2,g\left( 0 \right) = 0,f\left( 1 \right) = 6.. Let there exist a real number cc in [0,1] such that f′(c)=2g′(c),f'\left( c \right) = 2g'\left( c \right), then the value of g(1)g\left( 1 \right) must be
View answer and explanation
Correct option: B
2
Option B (2) is correct because by applying Rolle's theorem to the function ϕ(x)=f(x)−2g(x)\phi(x) = f(x) - 2g(x), we find that g(1) = 2.
33 Medium
33 of 281
Application +4 | -1
The point of the curve y2=2(x−3){y^2} = 2\left( {x - 3} \right) at which the normal is parallel to line   y−2x+1=0\;y - 2x + 1 = 0
View answer and explanation
Correct option: C
(5,  −2)\left( {5,\; - 2} \right)
Option C ((5, -2)) is correct because it is found by equating the slope of the normal to the slope of the given line.
34 Medium
34 of 281
Analyzing +4 | -1
Let PP be the point (other than the origin) of intersection of the curves y2=4ax{y^2} = 4ax and ay2=4x3a{y^2} = 4{x^3} such that the normals to the two curves meet xx-axis at G1 and G2{G_1}\,and\,{G_2} respectively. Then, G1G2={G_1}{G_2} =
View answer and explanation
Correct option: B
4a4a
Option B (4a4a) is correct because it is the distance between the x-intercepts of the normals to the two curves at their point of intersection.
35 Easy
35 of 281
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.
36 Easy
36 of 281
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.
37 Medium
37 of 281
Application +4 | -1
The line which is parallel to xx-axis and crosses the curve y=xy = \sqrt x an angle of 450{45^0}, is
View answer and explanation
Correct option: B
y=12y = \frac{1}{2}
Option B (y=12y = \frac{1}{2}) is correct because it represents the line parallel to the x-axis that intersects the curve at a 45-degree angle, determined by equating the derivative to tan(45).
38 Easy
38 of 281
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.
39 Easy
39 of 281
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.
40 Medium
40 of 281
Analyzing +4 | -1
Tangent of the angle at which the curves y=axy = {a^x} and y=bx  (a≠b>0)y = {b^x}\;(a \ne b > 0) intersect, is given by
View answer and explanation
Correct option: B
log⁡ab1+(log⁡a)(log⁡b)\frac{{\log \frac{a}{b}}}{{1 + \left( {\log a} \right)\left( {\log b} \right)}}
Option B (log⁡ab1+(log⁡a)(log⁡b)\frac{{\log \frac{a}{b}}}{{1 + \left( {\log a} \right)\left( {\log b} \right)}}) is correct because it is derived from the formula for the tangent of the angle between two curves using their derivatives at the point of intersection.
41 Medium
41 of 281
Application +4 | -1
Let f(x)=x3+x+1f(x)=x^3+x+1 and let g(x)g(x) be it's inverse function, then equation of the tangent to y=g(x)y=g(x) at x=3x=3 is
View answer and explanation
Correct option: A
x−4y+1=0x-4 y+1=0
Option A (x−4y+1=0x-4 y+1=0) is correct because it is the equation of the tangent line to the inverse function at the given point, found using the derivative of the inverse function.
42 Medium
42 of 281
Application +4 | -1
If x=t2x=t^2 and y=2ty=2 t, then the equation of the normal at t=1t=1 is
View answer and explanation
Correct option: A
x+y−3=0x+y-3=0
Option A (x+y−3=0x+y-3=0) is correct because by finding the derivative and the slope of the normal, we can determine the equation of the normal at the given point.
43 Easy
43 of 281
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.
44 Medium
44 of 281
Application +4 | -1
If the subnormal at any point on y=a1−nxny = {a^{1 - n}}{x^n} is of constant length, then the value of y=a1−nxny = {a^{1 - n}}{x^n}, is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because it makes the length of the subnormal constant, independent of x.
45 Medium
45 of 281
Analyzing +4 | -1
The tangent to the curve y=x3−6x2+9x+4,  0≤x≤5y = {x^3} - 6{x^2} + 9x + 4,\;0 \le x \le 5 has maximum slope at xx which is equal to
View answer and explanation
Correct option: D
None of these
Option C (Makes an acute angle with the xx-axis) is correct because the derivative is always positive, indicating an acute angle.

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46 Medium
46 of 281
Application +4 | -1
If ∫π/2x(3−2sin⁡2t)+∫0ycos⁡tdt=0\int\limits_{{\pi / 2}}^x \sqrt{\left(3-2 \sin ^2 t\right)}+\int\limits_{0}^y \cos t d t=0, then (dydx)π,π\left(\frac{d y}{d x}\right)_{\pi, \pi} is equal to
View answer and explanation
Correct option: C
3\sqrt{3}
Option C (3\sqrt{3}) is correct because after differentiating the given equation and substituting the given point, we find the derivative to be 3\sqrt{3}.
47 Easy
47 of 281
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.
48 Easy
48 of 281
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.
49 Easy
49 of 281
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.
50 Easy
50 of 281
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.
51 Easy
51 of 281
Application +4 | -1
The abscissa of the point on the curve

y=a(ex/a+e−x/a)y = a\left( {{e^{x/a}} + {e^{ - x/a}}} \right)

Where the tangent is parallel to the x-axis, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the derivative of the given function, representing the slope of the tangent, is zero at x=0, indicating a horizontal tangent.
52 Easy
52 of 281
Application +4 | -1
The equation of the tangent to the curve y=4e−−x4  y = 4{e^{ - - \frac{x}{4}}}\; at the point where the curve crosses yy-axis is equal to
View answer and explanation
Correct option: C
x+y=4x + y = 4
Option C (x+y=4x + y = 4) is correct because it represents the equation of the tangent line to the curve at the point where it crosses the y-axis, calculated using the derivative.
53 Medium
53 of 281
Application +4 | -1
The tangent to the curve y=ekxy=e^{k x} at a point (0,1)(0,1) meets the xx-axis at (a,0)(a, 0) where a∈[−2,−1]a \in[-2,-1], then k∈k \in :
View answer and explanation
Correct option: D
[12,1]\left[\frac{1}{2}, 1\right]
Option D ([12,1\frac{1}{2}, 1]) is correct because finding the tangent equation at (0,1) and using the given range for the x-intercept 'a' allows us to determine the range for k.
54 Medium
54 of 281
Application +4 | -1
The equations of those tangents to 4x2−9y2=364 x^2-9 y^2=36 which are perpendicular to the straight line 5x+2y−10=05 x+2 y-10=0, are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculation shows that there are no real points on the curve where the tangent is perpendicular to the given line.
55 Easy
55 of 281
Application +4 | -1
If x+4y=14x+4 y=14 is a normal to the curve y2=αx3−βy^{2}=\alpha x^{3}-\beta at (2,3)(2,3), then value of α+β\alpha+\beta is :
View answer and explanation
Correct option: A
9
Option A (9) is correct because substituting the point (2,3) into the curve equation and using the slope of the normal line allows us to solve for alpha and beta, resulting in alpha + beta = 9.
56 Easy
56 of 281
Analyzing +4 | -1
The curve x=y2 and xy=kx = {y^2}\,\mathrm{and}\,xy = k cut at right angles, if
View answer and explanation
Correct option: D
8k2=18{k^2} = 1
Option D (8k2=18k^2 = 1) is correct because it satisfies the condition for orthogonal intersection of the two curves, derived by equating the product of their slopes to -1.
57 Easy
57 of 281
Application +4 | -1
A normal to the hyperbola, 4x2−9y2=364 x^2-9 y^2=36 meets the coordinate axes xx and yy at AA and BB, respectively. If the parallelogram OABP(OO A B P(O being the origin) is formed, then the locus of PP is
View answer and explanation
Correct option: C
9x2−4y2=1699 x^2-4 y^2=169
Option C (9x2−4y2=1699 x^2-4 y^2=169) is correct because it represents the locus of point P derived from the properties of the normal to the hyperbola and the parallelogram OABP.
58 Easy
58 of 281
Analyzing +4 | -1
If OTOT and ONON are perpendiculars dropped from the origin to the tangent and normal to the curve x=asin⁡3tx=a \sin ^3 t, y=acos⁡3ty=a \cos ^3 t at an arbitrary point, then which one of the following is correct?
View answer and explanation
Correct option: D
All the above
Option D (All the above) is correct because all the given statements regarding the curve and its tangent/normal are true.
59 Medium
59 of 281
Analyzing +4 | -1
All points on the curve y2=4a(x+asin⁡xa){y^2} = 4a\left( {x + a\sin \frac{x}{a}} \right) at which the tangents are parallel to the axis of xx lie on a
View answer and explanation
Correct option: B
Parabola
Option B (Parabola) is correct because the points where the tangents are parallel to the x-axis lie on the parabola y2=4ax{y^2} = 4ax.
60 Easy
60 of 281
Application +4 | -1
For the curve x=t2−1,y=t2−tx=t^2-1, y=t^2-t, the tangent line is perpendicular to xx-axis when
View answer and explanation
Correct option: A
t=0t=0
The correct answer is t=0t=0 because at this value, the tangent line to the curve is perpendicular to the x-axis, meaning dxdy=0\frac{dx}{dy} = 0.
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