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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Mixed difficulty

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Showing 121-150 of 281 questions

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121 Medium
121 of 281
Application +4 | -1
The equation of tangent to the curve y=be−x/ay = b{e^{ - x/a}} 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 C (0.5) is correct because it represents the limit of θ as h approaches 0, derived using the Mean Value Theorem.
122 Medium
122 of 281
Application +4 | -1
Let P(2,  2)P\left( {2,\;2} \right) and Q(1/2,  −1)Q\left( {1/2,\; - 1} \right) be two points on the parabola y2=2x{y^2} = 2x. The coordinates of the point RR on the parabola y2=2x{y^2} = 2x, where the tangent to the curve is parallel to the chord PQPQ, are
View answer and explanation
Correct option: B
(1/8,  1/2)\left( {1/8,\;1/2} \right)
Option B ((1/8,  1/2)\left( {1/8,\;1/2} \right)) is correct because it represents the point on the parabola where the tangent is parallel to the chord PQ.
123 Medium
123 of 281
Application +4 | -1
The line x+y=2x+y=2 is tangent to the curve x2=3−2yx^2=3-2 y at its point
View answer and explanation
Correct option: A
(1,1)(1,1)
Option A ((1,1)(1,1)) is correct because by finding the derivative of the curve and setting it equal to the slope of the line, we can solve for the point of tangency.
124 Easy
124 of 281
Application +4 | -1
The point on the curve y2=xy^2=x where tangent makes 45∘45^{\circ} angle with xx-axis, is
View answer and explanation
Correct option: B
(1/4   1/2)(1 / 4 \,\,\,1 / 2)
The correct answer is B ((1/4  1/2)(1 / 4 \,\,1 / 2)) because at this point on the curve, the slope of the tangent is 1, which corresponds to a 45-degree angle with the x-axis.
125 Medium
125 of 281
Application +4 | -1
If f(x)=log⁡xf(x)=\log x satisfies Lagrange's theorem on [1,e][1, e] then value of c∈(1,e)c \in(1, e) such that the tangent at cc is parallel to line joining (1,f(1))(1, f(1)) and (e,f(e))(e, f(e)) is
View answer and explanation
Correct option: C
e−1e-1
Option C (e−1e-1) is correct because the value of c that satisfies Lagrange's theorem is e-1, which is found by equating the derivative at c to the average rate of change over the interval [1, e].
126 Medium
126 of 281
Application +4 | -1
Equation of normal to the curve x2=4yx^2=4 y which passes through the point (1,2)(1,2) is
View answer and explanation
Correct option: B
x+y=3x+y=3
Option B x+y=3x+y=3 is correct because by finding the equation of the normal to the curve and using the given point, we can solve for the normal's equation.
127 Easy
127 of 281
Application +4 | -1
The length of the subtangent at (2,2) to the curve x5=2y4  {x^5} = 2{y^4}\; is
View answer and explanation
Correct option: B
85\frac{8}{5}
Option B (85\frac{8}{5}) is correct because it represents the length of the subtangent calculated using the derivative of the curve at the given point.
128 Easy
128 of 281
Application +4 | -1
If curve represented by y=(x−a)(x−3)(x−2)y=\frac{(x-a)}{(x-3)(x-2)} passes through (1,–3)(1, –3) then equation of normal at (1,–3)(1, –3) to
the curve is given by
View answer and explanation
Correct option: D
x−4y=13x-4y=13
Option D (x - 4y = 13) is correct because it represents the equation of the normal to the curve at the point (1, -3), calculated using the derivative of the function and the point-slope form of a line.
129 Easy
129 of 281
Application +4 | -1
If the curves 2x2+3y2=62 x^2+3 y^2=6 and ax2+4y2=4a x^2+4 y^2=4 intersect orthogonally, then a=a=
View answer and explanation
Correct option: A
2
Option A (2) is correct because the value of 'a' for the curves to intersect orthogonally is calculated to be 2.
130 Easy
130 of 281
Application +4 | -1
The normal to the curve y(x−2)(x−3)=x+6y(x-2)(x-3)=x+6 at the point where the curve intersects the yy-axis passes through the point
View answer and explanation
Correct option: B
(12,12)\left(\frac{1}{2}, \frac{1}{2}\right)
Option B ((1/2, 1/2)) is correct because the normal to the curve at (0,1) passes through the point (1/2, 1/2).
131 Easy
131 of 281
Analyzing +4 | -1
If θ\theta is the angle between the curves x  y=2x\;y = 2 and x2+4y=0{x^2} + 4y = 0, then tan⁡θ\tan \theta is equal to
View answer and explanation
Correct option: D
3
Option D (3) is correct because it represents the tangent of the angle between the two curves at their point of intersection, calculated using their derivatives.
132 Medium
132 of 281
Analyzing +4 | -1
The angle of intersection of the curves y=x2 and x=y2y = {x^2}\,and\,x = {y^2} is
View answer and explanation
Correct option: D
tan⁡−1(34){\tan ^{ - 1}}\left( {\frac{3}{4}} \right)
Option D (tan⁡−1(34){\tan ^{ - 1}}(\frac{3}{4})) is correct because it is the angle calculated using the slopes of the two curves at their point of intersection.
133 Easy
133 of 281
Analyzing +4 | -1
If sin⁡−1θ\sin ^{-1} \theta is the acute angle between the curves x2+y2=4xx^2+y^2=4 x and x2+y2=8x^2+y^2=8 at (2,2)(2,2), then θ=\theta=
View answer and explanation
Correct option: C
1/21 / \sqrt{2}
Option C (1/21 / \sqrt{2}) is correct because the angle between the curves at the intersection point is π/4\pi/4, and sin⁡−1(1/2)=π/4\sin^{-1}(1/\sqrt{2}) = \pi/4.
134 Easy
134 of 281
Analyzing +4 | -1
If x2a2+y2b2=1(a>b)\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1(a > b) and x2−y2=c2{x^2} - {y^2} = {c^2} cut at right angles, then
View answer and explanation
Correct option: C
a2−b2=2c2{a^2} - {b^2} = 2{c^2}
Option C (a2−b2=2c2{a^2} - {b^2} = 2{c^2}) is correct because it satisfies the condition for orthogonal intersection of the ellipse and hyperbola, derived by equating the product of their slopes to -1.
135 Easy
135 of 281
Application +4 | -1
The value of mm for which the area of the triangle included between the axes and any tangent to the curve xmy=bmx^m y=b^m is constant, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the area of the triangle formed by the tangent and the axes is constant only when m=1.

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136 Easy
136 of 281
Analyzing +4 | -1
If the line ax+by+c=0a x+b y+c=0 is a tangent to the curve xy=9x y=9, then
View answer and explanation
Correct option: A
a>0,b>0a > 0, b > 0
Option A (a>0,b>0a > 0, b > 0) is correct because the signs of a and b must be the same for the line to be tangent to the curve.
137 Hard
137 of 281
Application +4 | -1
Let x2+y2+Ax+By+C=0{x^2} + {y^2} + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola y=x2y = {x^2} at (2, 4). Then A+CA + C is equal to ______.
View answer and explanation
Correct option: A
16
Option B (96) is correct because it is the solution to the system of equations derived from the given information about the arithmetic progression.
138 Easy
138 of 281
Application +4 | -1
The point on the curve y=x3y = {x^3} at which the tangent to the curve is parallel to the xx-axis, is
View answer and explanation
Correct option: D
(0,0)
Option D ((0,0)) is correct because the tangent is parallel to the x-axis when the derivative dy/dx is zero, which occurs at x=0 for the given curve.
139 Medium
139 of 281
Application +4 | -1
The curve y−exy+x=0y - {e^{xy}} + x = 0 has a vertical tangent at the point
View answer and explanation
Correct option: A
(1, 0)
Option A ((1, 0)) is correct because it satisfies the condition for a vertical tangent, where the derivative approaches infinity.
140 Easy
140 of 281
Application +4 | -1
The tangent to the curve y−ax2+bxy-a x^2+b x at (2,−8)(2,-8) is parallel to xx-axis. Then
View answer and explanation
Correct option: C
a=2,b=−8a=2, b=-8
Option C (a=2, b=-8) is correct because these values satisfy both the condition that the tangent at (2, -8) is parallel to the x-axis (dy/dx = 0) and that the point lies on the curve.
141 Medium
141 of 281
Application +4 | -1
The length of the normal to the curve x=a(θ+sin⁡θ),y=a(1−cos⁡θ)at  θ=π2x = a\left( {\theta + \sin \theta } \right),y = a\left( {1 - \cos \theta } \right)at\;\theta = \frac{\pi }{2} is
View answer and explanation
Correct option: D
2a\sqrt {2} a
Option D (2a\sqrt {2} a) is correct because it is the calculated length of the normal to the curve at the given parameter value.
142 Easy
142 of 281
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.
143 Medium
143 of 281
Application +4 | -1
The value of n\mathrm{n} for which the length of subnormal of the curve xyn=an+1x y^n=a^{n+1} is constant, is
View answer and explanation
Correct option: D
-2
Option D (-2) is correct because it makes the length of the subnormal constant, independent of x and y.
144 Medium
144 of 281
Application +4 | -1
The angle between the curves y=x2 and y2−x=0y = {x^2}\,and\,{y^2} - x = 0 at the point (1,1), is
View answer and explanation
Correct option: D
tan⁡−134{\tan ^{ - 1}}\frac{3}{4}
Option D (tan⁡−134{\tan ^{ - 1}}\frac{3}{4}) is correct because it represents the angle between the two curves at their point of intersection (1,1), calculated using the slopes of their tangents.
145 Hard
145 of 281
Application +4 | -1
If the area of the triangle, included between the axes and any tangent to the curve xyn=an+1x{y^n} = {a^{n + 1}} is constant, then the value of nn is
View answer and explanation
Correct option: C
1
Option C (1) is correct because when n=1n=1, the area of the triangle formed by the tangent is constant.
146 Easy
146 of 281
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.
147 Easy
147 of 281
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.
148 Medium
148 of 281
Application +4 | -1
For the parabola y2=4ax{y^2} = 4ax, the ratio of the subtangent to the abscissa is
View answer and explanation
Correct option: B
2:12:1
Option B (x+2;y=π2x + 2\\;y = \frac{\pi }{2}) is correct because it is the equation of the tangent to the curve that is parallel to the given line.
149 Easy
149 of 281
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.
150 Easy
150 of 281
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.
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