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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Mixed difficulty

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Showing 241-270 of 281 questions

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241 Easy
241 of 281
Application +4 | -1
If the tangent to the curve, y=x3+ax−b y = {x^3} + ax - b\, at the point (1, -5) is perpendicular to the line, −x+y+4=0, - x + y + 4 = 0,\, then which one of the following points lies on the curve?
View answer and explanation
Correct option: B
(2, -2)
Option B ((2, -2)) is correct because, after finding the values of a and b using the tangent condition and the point (1, -5), the point (2, -2) satisfies the curve equation.
242 Medium
242 of 281
Application +4 | -1
The point at which the tangent to the curve y=2x2−x+1y = 2{x^2} - x + 1 is parallel to y=3x+9y = 3x + 9, will be
View answer and explanation
Correct option: B
(1,2)
Option B (1,2) is correct because the tangent to the curve is parallel to the line when their slopes are equal, which occurs at the point (1, 2).
243 Medium
243 of 281
Application +4 | -1
Let ll be the line through (0,0)(0,0) and tangent to the curve y=x3+x+16y=x^{3}+x+16. Then the slope of ll equal to :
View answer and explanation
Correct option: D
13
Option D (13) is correct because finding the tangent line equation and using the point of tangency allows us to solve for the slope 'm', which equals 13.
244 Easy
244 of 281
Application +4 | -1
The point at which the tangent to the curve y=x2−4xy=x^2-4 x is parallel to xx-axis, is
View answer and explanation
Correct option: D
(2,−4)(2,-4)
Option D ((2,-4)) is correct because at this point the derivative of the curve is zero, indicating a horizontal tangent parallel to the x-axis.
245 Medium
245 of 281
Analyzing +4 | -1
If the curves x2a2+y212=1\frac{x^2}{a^2}+\frac{y^2}{12}=1 and y3=8xy^3=8 x intersect at right angles, then the value of a2a^2 is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because it is the value of a2a^2 that satisfies the condition for the curves to intersect at right angles.
246 Medium
246 of 281
Analyzing +4 | -1
The maximum slope of the curve y=−x3+3x2+2x−27y = - {x^3} + 3{x^2} + 2x - 27 is
View answer and explanation
Correct option: A
5
Option A (5) is correct because it is the maximum value of the derivative of the given function, found by setting the second derivative to zero.
247 Easy
247 of 281
Application +4 | -1
The slope of the normal to the curve y=2x2+3sin⁡xy=2 x^2+3 \sin x at x=0x=0 is
View answer and explanation
Correct option: D
−13-\frac{1}{3}
Option D (−13-\frac{1}{3}) is correct because it is the negative reciprocal of the slope of the tangent to the curve at x=0, which represents the slope of the normal.
248 Medium
248 of 281
Analyzing +4 | -1
The shortest distance between the line y−x=1y - x = 1 and the curve x=y2x = {y^2} is
View answer and explanation
Correct option: A
328\frac{{3\sqrt 2 }}{8}
Option A (328\frac{{3\sqrt 2 }}{8}) is correct because it represents the shortest distance between the line and the curve, calculated using the formula for the distance between a point and a line.
249 Easy
249 of 281
Application +4 | -1
The equation of tangent to the curve x23−y22=1\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1, which is parallel to y=xy = x, is
View answer and explanation
Correct option: A
y=x±1y = x \pm 1
Option A (y=x±1y = x \pm 1) is correct because it represents the equations of the tangents to the hyperbola that are parallel to the line y=x.
250 Easy
250 of 281
Application +4 | -1
The slope of normal at any point (x,y),x>0,y>0(x,y),x > 0,y > 0 on the curve y=y(x)y = y(x) is given by x2xy−x2y2−1.\frac{{{x^2}}}{{xy - {x^2}{y^2} - 1}}. If the curve passes through the point (1, 1) then e.y(e)e.y(e) is equal to
View answer and explanation
Correct option: D
1+tan⁡(1)1−tan⁡(1)\frac{{1 + \tan (1)}}{{1 - \tan (1)}}
Option D (1+tan⁡(1)1−tan⁡(1)\frac{{1 + \tan (1)}}{{1 - \tan (1)}}) is correct because by solving the differential equation and applying the initial condition, we can find the value of e.y(e)e.y(e).
251 Easy
251 of 281
Application +4 | -1
The equation of the tangent to the curve (1+x2)y=2−x\left( {1 + {x^2}} \right)y = 2 - x, where it crosses the xx-axis, is
View answer and explanation
Correct option: A
x+5y=2x + 5y = 2
Option A (x+5y=2x + 5y = 2) is correct because it represents the equation of the tangent line to the curve at the point where it crosses the x-axis, calculated using derivatives.
252 Easy
252 of 281
Application +4 | -1
If y=4x−5y = 4x - 5 is a tangent to the curve y2=px3+q{y^2} = p{x^3} + q at (2, 3) then
View answer and explanation
Correct option: A
p=2,  q=−7p = 2,\;q = - 7
Option A (p=2,  q=−7p = 2,\;q = - 7) is correct because these values of p and q satisfy both the equation of the curve at the given point and the slope of the tangent at that point.
253 Medium
253 of 281
Application +4 | -1
The equation of the tangent to the curve y=9−2x2y=\sqrt{9-2 x^2} at the point where the ordinate & the abscissa are equal is
View answer and explanation
Correct option: D
2x+y−33=02 x+y-3 \sqrt{3}=0
Option D (2x+y−33=02 x+y-3 \sqrt{{3}}=0) is correct because it is the equation of the tangent calculated at the point where the ordinate and abscissa are equal on the curve.
254 Easy
254 of 281
Application +4 | -1
The tangent to the curve y=xex2y=x e^{x^2} passing through the point (1,e)(1, e) also passes through the point
View answer and explanation
Correct option: A
(43,2e)\left(\frac{4}{3}, 2 e\right)
Option A (43,2e)\left(\frac{4}{3}, 2 e\right) is correct because it satisfies the equation of the tangent line to the curve at the point (1, e).
255 Easy
255 of 281
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.

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256 Hard
256 of 281
Application +4 | -1
The slope of the tangent to the curve y=∫xx2cos⁡−1t2dty=\int\limits_{x}^{x^2} \cos ^{-1} t^2 d t at x=2−1/4x=2^{-1 / 4}, is
View answer and explanation
Correct option: B
(843−14)π\left(\frac{\sqrt[4]{8}}{3}-\frac{1}{4}\right) \pi
Option B ((843−14)π\left(\frac{\sqrt[4]{8}}{3}-\frac{1}{4}\right) \pi) is correct because the slope of the tangent is found by differentiating the given integral using Leibniz's rule and evaluating it at x=2−1/4x=2^{-1/4}.
257 Medium
257 of 281
Application +4 | -1
The length of subtangent to the curve x2y2=a4{x^2}{y^2} = {a^4} at the point (−a,  a)\left( { - a,\;a} \right) is
View answer and explanation
Correct option: C
aa
Option C (aa) is correct because the length of the subtangent is calculated by finding the derivative and using the formula for subtangent length.
258 Medium
258 of 281
Application +4 | -1
The equation(s) of the tangent(s) to the curve y=x4y = {x^4} from the point (2, 0) not on the curve is given by
View answer and explanation
Correct option: C
y−409681=204827(x−83)y - \frac{{4096}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)
Option C (y−409681=204827(x−83)y - \frac{{4096}}{{81}} = \frac{{2048}}{{27}}\left( {x - \frac{8}{3}} \right)) is correct because it represents the tangent to the curve y=x4y = {x^4} from the point (2, 0) not on the curve.
259 Easy
259 of 281
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.
260 Easy
260 of 281
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.
261 Medium
261 of 281
Application +4 | -1
The point on the curve 3y=6x−5x33 y=6 x-5 x^3 the normal at which passes through the origin, is
View answer and explanation
Correct option: A
(1,1/3)(1,1 / 3)
Option A ((1, 1/3)) is correct because the normal to the curve at this point passes through the origin, satisfying both the curve's equation and the normal's equation.
262 Medium
262 of 281
Application +4 | -1
The length of subnormal of parabola y2=4ax{y^2} = 4ax at any point is equal to
View answer and explanation
Correct option: D
2a2a
Option D ((32,  −172)\left( {\frac{3}{2},\; - \frac{{17}}{2}} \right)) is correct because it is the point where the derivative of the function is zero, indicating a horizontal tangent.
263 Medium
263 of 281
Application +4 | -1
If tangent to the curve x=at2,  y=2atx = a{t^2},\;y = 2at is perpandicular to xx-axis, then its point of contact is
View answer and explanation
Correct option: C
(0,0)
Option D (a=11a = 11) is correct because it satisfies the conditions of Rolle's theorem and the given derivative value.
264 Easy
264 of 281
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.
265 Easy
265 of 281
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.
266 Easy
266 of 281
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.
267 Easy
267 of 281
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.
268 Easy
268 of 281
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.
269 Easy
269 of 281
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.
270 Hard
270 of 281
Analyzing +4 | -1
The points of contact of the tangents drawn from the origin to the curve y=sin⁡xy = \sin x lie on the curve
View answer and explanation
Correct option: C
x2−y2=x2y2{x^2} - {y^2} = {x^2}{y^2}
The correct answer is C (x2−y2=x2y2{x^2} - {y^2} = {x^2}{y^2}) because it represents the locus of the points of contact of tangents drawn from the origin to the curve y=sin⁡xy = \sin x.
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