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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Easy

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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 91-120 of 164 questions

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91 Easy
91 of 164
Application +4 | -1
The length of the normal to the curve y=a(e−x/a+ex/a2)y=a\left(\frac{e^{-x / a}+e^{x / a}}{2}\right) at any point varies as the
View answer and explanation
Correct option: D
Square of the ordinate of the point
Option D (Square of the ordinate of the point) is correct because the length of the normal varies as the square of the ordinate.
92 Easy
92 of 164
Application +4 | -1
The abscissae of the point on the curve xy=(a+x)2x y=(a+x)^2, the normal at which cuts off numerically equal intercepts from the axes of coordinates, is
View answer and explanation
Correct option: A
a2\frac{a}{\sqrt{2}}
Option A (a2\frac{a}{\sqrt{2}}) is correct because the abscissa of the point is found by setting the negative reciprocal of the derivative equal to ±1\pm 1 and solving for xx.
93 Easy
93 of 164
Application +4 | -1
Let g(x)=  {2e                          if  x≤1  log(x−1),      if    x>1  g\left( x \right) = \;\left\{ {\begin{array}{cccccccccccccccccccc} {2e\;\;\;\;\;\;\;\;\;\;\;\;\;if\;x \le 1}\\ {\mathrm{\;log}\left( {x - 1} \right),\;\;\;if\;\;x > 1} \end{array}} \right.\; . The equation of the normal to y=g(x)y = g\left( x \right) at the point (3,log 2), is
View answer and explanation
Correct option: B
y+2x=6+log⁡2y + 2x = 6 + \log 2
Option B (y+2x=6+log⁡2y + 2x = 6 + \log 2) is correct because it represents the equation of the normal line to the curve at the given point, calculated using the derivative.
94 Easy
94 of 164
Application +4 | -1
If the normal to the curve y=f(x)y=f(x) at the point (3,4)(3,4) makes an angle 3π4\frac{3 \pi}{4} with the positive xx-axis, then f′(3)=f^{\prime}(3)=
View answer and explanation
Correct option: D
1
Option D (1) is correct because the derivative of the function at the given point is equal to the slope of the tangent, and the normal's slope is the negative reciprocal.
95 Easy
95 of 164
Application +4 | -1
If the slope of the curve y=axb−xy=\frac{a x}{b-x} at the point (1,1)(1,1) is 2, then
View answer and explanation
Correct option: C
a=1,b=2a=1, b=2
Option C (a=1,b=2a=1, b=2) is correct because substituting the point (1,1) and the slope 2 into the derivative of the curve equation yields these values for a and b.
96 Easy
96 of 164
Application +4 | -1
The equation of the tangent to the curve y=1−ex/2y=1-e^{x / 2} at the point of intersection with the yy-axis, is
View answer and explanation
Correct option: A
x+2y=0x+2 y=0
Option A (x+2y=0x+2y=0) is correct because it represents the equation of the tangent line to the curve at the point (0,0) with a slope of -1/2.
97 Easy
97 of 164
Application +4 | -1
The equation of the tangent to the curve y=e−∣x∣y = {e^{ - \left| x \right|}} at the point where the curve cuts the line x=1x = 1 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated equation of the tangent does not match any of the given options.
98 Easy
98 of 164
Analyzing +4 | -1
If the slope of the tangent to the curve y=excos⁡xy = {e^x}\cos x is minimum at x=a,  0≤a≤2  πx = a,\;0 \le a \le 2\;\pi, then the value of aa is
View answer and explanation
Correct option: B
π\pi
Option B (π\pi) is correct because the second derivative of the slope is positive at x=πx = \pi, indicating a minimum slope.
99 Easy
99 of 164
Application +4 | -1
The length of tangent, subtangent, normal and subnormal for the curve y=x2+x−1y = {x^2} + x - 1 at (1, 1) are A,  B,  C and DA,\;B,\;C\,\mathrm{and}\,D respectively, then their increasing order is
View answer and explanation
Correct option: D
B,  A,  D,  CB,\;A,\;D,\;C
Option B (2 units) is correct because it is the length of the side when the rate of change of area equals the rate of change of perimeter.
100 Easy
100 of 164
Application +4 | -1
Equation of a common tangent (if it exists) to x2+y2=2x^2+y^2=2 and x3y3=1x^3 y^3=1 is
View answer and explanation
Correct option: C
x+y=2x+y=2
Option C (x+y=2x+y=2) is correct because it represents a common tangent to both the circle and the curve, found by equating the derivatives and solving for intersection points.
101 Easy
101 of 164
Application +4 | -1
If STST and SNSN are the lengths of the subtangent and the subnormal at the point θ=π2\theta = \frac{\pi }{2} on the curve x=a(θ+sin⁡θ),  y=a(1−cos⁡θ),  a≠1x = a\left( {\theta + \sin \theta } \right),\;y = a\left( {1 - \cos \theta } \right),\;a \ne 1, then
View answer and explanation
Correct option: A
ST=SNST = SN
Option A (ST=SNST = SN) is correct because at θ=π2\theta = \frac{\pi}{2}, the lengths of the subtangent and subnormal are both equal to aa.
102 Easy
102 of 164
Application +4 | -1
The equation of tangent at (−4,−4)(-4,-4) on the curve x2=−4yx^2=-4 y is
View answer and explanation
Correct option: D
2x−y+4=02 x-y+4=0
Option D (2x−y+4=02 x-y+4=0) is the correct answer because it represents the equation of the tangent line to the curve at the given point.
103 Easy
103 of 164
Application +4 | -1
The slope of the tangent to the curve y=∫0xdx1+x3\mathrm{y}=\int_0^{\mathrm{x}} \frac{\mathrm{dx}}{1+\mathrm{x}^3} at the point where x=1\mathrm{x}=1 is -
View answer and explanation
Correct option: A
1/21 / 2
Option A (1/2) is correct because the derivative of the integral gives 11+x3\frac{{1}}{{1+x^3}}, and evaluating at x=1 yields 1/2.
104 Easy
104 of 164
Application +4 | -1
The equation of normal to the curve x2y=x2−3x+6{x^2}y = {x^2} - 3x + 6 at the point with abscissa x=3x = 3 is
View answer and explanation
Correct option: B
27x−3y=7927x - 3y = 79
Option B (27x−3y=7927x - 3y = 79) is correct because it represents the equation of the normal to the curve at the specified point, calculated using the derivative and the negative reciprocal slope.
105 Easy
105 of 164
Analyzing +4 | -1
The intercept on xx-axis made by tangents to the curve y=∫0x∣t∣dt,x∈Ry=\int\limits_{0}^x|t| d t, x \in R, which are parallel to the line y=2xy=2 x, are equal to
View answer and explanation
Correct option: A
±1\pm 1
Option A (±1\pm 1) is correct because the x-intercepts of the tangents parallel to y=2x are ±1\pm 1.
106 Easy
106 of 164
Application +4 | -1
If an equation of a tangent to the curve, y−cos⁡(x+y),−1−1≤x≤1+πy-\cos (x+y),-1-1 \leq x \leq 1+\pi, is x+2y=kx+2 y=k then kk is equal to :
View answer and explanation
Correct option: D
π2\frac{\pi}{2}
Option D (π2\frac{{\pi}}{{2}}) is correct because it is the value of k obtained by finding the point on the curve where the tangent has the given slope and substituting it into the tangent equation.
107 Easy
107 of 164
Analyzing +4 | -1
The angle between the curves xy=2x y=2 and y2=4xy^2=4 x at their point of intersection is
View answer and explanation
Correct option: B
tan⁡−13\tan ^{-1} 3
The correct answer is B (tan⁡−13\tan^{-1} 3) because by finding the slopes of the tangents at the point of intersection and using the formula for the angle between two lines, the angle is tan⁡−13\tan^{-1} 3.
108 Easy
108 of 164
Application +4 | -1
The points at which the tangents to the curve ax2+a x^2+ 2hxy+by2=12 h x y+b y^2=1 is parallel to yy-axis is
View answer and explanation
Correct option: B
where hx+by=0h x+b y=0 meets it
Option B (where hx+by=0hx+by=0 meets it) is correct because the tangent is parallel to the y-axis when the denominator of the derivative is zero.
109 Easy
109 of 164
Application +4 | -1
If the tangent at any point on the curve x4+y4=c4x^4+y^4=c^4 cuts off intercepts aa and bb on the coordinate axes, the value of a−4/3+b−4/3a^{-4 / 3}+b^{-4 / 3} is
View answer and explanation
Correct option: A
c−4/3c^{-4 / 3}
Option A (c−4/3c^{-4 / 3}) is correct because the equation of the tangent and the given curve are used to find the intercepts and simplify the expression.
110 Easy
110 of 164
Application +4 | -1
The length of the normal to the curve x=a(θ+sin⁡θ),y=a(1−cos⁡θ)x=a(\theta+\sin \theta), y=a(1-\cos \theta), at θ=π/2\theta=\pi / 2, is
View answer and explanation
Correct option: B
a2a \sqrt{2}
Option B (a2a \sqrt{2}) is correct because the length of the normal to the curve at θ=π/2\theta = \pi/2 is calculated to be a2a\sqrt{2}.
111 Easy
111 of 164
Analyzing +4 | -1
Any tangent to the curve y=2x5+4x3+7x+9y = 2{x^5} + 4{x^3} + 7x + 9
View answer and explanation
Correct option: C
Makes an acute angle with the xx-axis
Option A (Parallel) is correct because the tangents at the points where the curve crosses the x-axis have the same slope.
112 Easy
112 of 164
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
Option A (t=0) is correct because the tangent line is perpendicular to the x-axis when dx/dt = 0 and dy/dt is not 0.
113 Easy
113 of 164
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.
114 Easy
114 of 164
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.
115 Easy
115 of 164
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.
116 Easy
116 of 164
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.
117 Easy
117 of 164
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.
118 Easy
118 of 164
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.
119 Easy
119 of 164
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.
120 Easy
120 of 164
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.
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