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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 151-164 of 164 questions

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151 Easy
151 of 164
Understanding +4 | -1
At any point of a curve (subtangent) ×\times (subnormal) is equal to the square of the
View answer and explanation
Correct option: D
Ordinate of that point
Option A (1) is correct because as x approaches 4 from the right, (x-4) is positive, so the absolute value simplifies to (x-4), and the fraction equals 1.
152 Easy
152 of 164
Analyzing +4 | -1
A tangent to the curve y=∫0x∣x∣dty=\int\limits_{0}^x|x| d t, which is parallel to the line y=xy=x, cuts off an intercept from the yy-axis is equal to
View answer and explanation
Correct option: B
−12,12\frac{-1}{2}, \frac{1}{2}
Option B (−12,12\frac{{-1}}{{2}}, \frac{{1}}{{2}}) is correct because it represents the y-intercepts of the tangent lines parallel to y=x.
153 Easy
153 of 164
Application +4 | -1
The point on the curve y=12x−x2y=12 x-x^2 where the tangent is parallel to xx-axis, is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because the point where the tangent is parallel to the x-axis is (6, 36), which is not among the options.
154 Easy
154 of 164
Application +4 | -1
The curve y=ax3+bx2+cx+8y=a x^3+b x^2+c x+8 touches xx-axis at P(−2,0)P(-2,0) and cuts the yy-axis at point QQ where its gradient is 3 . The values of a,b,ca, b, c are respectively
View answer and explanation
Correct option: D
−1/4,0,3-1 / 4,0,3
Option D (-1/4, 0, 3) is correct because these values satisfy the given conditions of the curve touching the x-axis and having a specific gradient at the y-axis intercept.
155 Easy
155 of 164
Application +4 | -1
If y=f(x)y=f(x) be the equation of the line touching the line y=2x+3y=2 x+3 at x=2x=2, then
View answer and explanation
Correct option: B
2f(2)=7f′(2)2 f(2)=7 f^{\prime}(2)
Option B (2f(2)=7f′(2)2 f(2)=7 f^{\prime}(2)) is correct because, since the curve touches the line at x=2, the derivative of f(x) at x=2 is equal to the slope of the line, and f(2) is equal to the y-value of the line at x=2.
156 Easy
156 of 164
Application +4 | -1
The length of the subtangent at any point (x1,  y1)\left( {{x_1},\;{y_1}} \right) on the curve y=ax,  (a>0)y = {a^x},\;(a > 0) is
View answer and explanation
Correct option: B
1log⁡a\frac{1}{{\log a}}
Option B (1log⁡a\frac{1}{{\log a}}) is correct because it represents the length of the subtangent calculated using the derivative of the curve and the formula for subtangent length.
157 Easy
157 of 164
Application +4 | -1
The length of the tangent to the curve x=asin⁡3t,y=acos⁡3t(a>0)x = a{\sin ^3}t,y = a{\cos ^3}t(a > 0) at any arbitrary is
View answer and explanation
Correct option: A
acos⁡2ta{\cos ^2}t
Option A (acos⁡2ta{\cos ^2}t) is correct because it's the length of the tangent calculated using derivatives and the formula for tangent length.
158 Easy
158 of 164
Application +4 | -1
The angle between the curves y=sin⁡x and   y=cos⁡xy = \sin x\,and\,\;y = \cos x is
View answer and explanation
Correct option: A
tan⁡−1(22){\tan ^{ - 1}}\left( {2\sqrt 2 } \right)
Option A (tan⁡−1(22){\tan ^{ - 1}}( {2\sqrt 2 } )) is correct because it represents the angle between the two curves, calculated using the slopes of the tangent lines at their intersection point.
159 Easy
159 of 164
Application +4 | -1
The normal to the curve x2+2xy−3y2=0x^2+2 x y-3 y^2=0, at (1,1)(1,1):
View answer and explanation
Correct option: B
Meets the curve again the fourth quadrant.
The correct answer is 'Meets the curve again the fourth quadrant' because the normal at (1,1) intersects the curve again at (3,-1), which lies in the fourth quadrant.
160 Easy
160 of 164
Application +4 | -1
If mm is the slope of the tangent to the curve ey=1+x2e^y=1+x^2, then
View answer and explanation
Correct option: D
∣m∣≤1|m| \leq 1
Option D (∣m∣≤1|m| \leq 1) is correct because the slope of the tangent, ∣m∣|m|, is always less than or equal to 1, as derived using AM-GM inequality.
161 Easy
161 of 164
Analyzing +4 | -1
The two curves x3−3xy2+2=0x^3-3 x y^2+2=0 and 3x2y−y3=23 x^2 y-y^3=2
View answer and explanation
Correct option: A
Cut at right angles
Option A (Cut at right angles) is correct because the product of the slopes of the tangents to the two curves at their point of intersection is -1.
162 Easy
162 of 164
Analyzing +4 | -1
At which point the line xa+yb=1\frac{x}{a}+\frac{y}{b}=1, touches the curve y=be−xiay=b e^{-x i a}
View answer and explanation
Correct option: C
(0,b)(0, b)
Option C ((0, b)) is correct because it is the point where the line and the curve have the same slope, indicating tangency.
163 Easy
163 of 164
Application +4 | -1
The point on the curve y=(x−1)(x−2)y=(x-1)(x-2) at which the tangent makes an angle of 135∘135^{\circ} with the positive direction of xx-axis has coordinates
View answer and explanation
Correct option: A
(1,0)(1,0)
Option A ((1,0)) is correct because at this point, the derivative of the curve equals -1, which corresponds to a tangent angle of 135 degrees with the x-axis.
164 Easy
164 of 164
Analyzing +4 | -1
If the parabolas y2=4x{y^2} = 4x and x2=32y{x^2} = 32y intersect at (16,8)\left( {16,8} \right) at an angle θ,\theta , then θ\theta is equal to
View answer and explanation
Correct option: A
tan⁡−1(3/5){\tan ^{ - 1}}\left( {3/5} \right)
Option A (tan⁡−1(3/5){\tan ^{ - 1}}\left( {3/5} \right)) is correct because it is the angle between the tangents to the parabolas at the point of intersection, calculated using the slopes of the tangents.
Revision notes and key takeaways

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