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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Easy

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Showing 61-90 of 164 questions

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61 Easy
61 of 164
Application +4 | -1
A tangent to a curve intersects the y\mathrm{y}-axis at a point P\mathrm{P}. A line perpendicular to this tangent and passing through P\mathrm{P} also passes through (1,0)(1,0). The differential equation of the curve is
View answer and explanation
Correct option: A
yy′−x(y′)2−1y y^{\prime}-x\left(y^{\prime}\right)^2-1
Option A is correct because it represents the differential equation derived from the given geometric conditions involving tangents and perpendicular lines.
62 Easy
62 of 164
Application +4 | -1
The area bounded by the coordinate axes and normal to the curve y=log⁡exy=\log _e x at the point P(1,0)P(1,0), is
View answer and explanation
Correct option: C
12\frac{1}{2} sq. unit
Option C (1/2 sq. unit) is correct because the area bounded by the coordinate axes and the normal is calculated to be 1/2 square unit.
63 Easy
63 of 164
Application +4 | -1
The line (x/a)+(y/b)=2\left( {x/a} \right) + \left( {y/b} \right) = 2 touches the curvexnan+ynbn=2\frac{{{x^n}}}{{{a^n}}} + \frac{{{y^n}}}{{{b^n}}} = 2, at
View answer and explanation
Correct option: C
(a,  b)\left( {a,\;b} \right)
Option C (a,  b)\left( {a,\;b} \right) is the correct answer because it satisfies both the line and curve equations at the point of tangency.
64 Easy
64 of 164
Application +4 | -1
The slope of the tangent to the curves x=t2+3t−8,  y=2t2−2t−5x = {t^2} + 3t - 8,\;y = 2{t^2} - 2t - 5 at the point (2,  −1)  (2,\; - 1)\; is
View answer and explanation
Correct option: B
67\frac{6}{7}
Option B (67\frac{6}{7}) is correct because it is the slope of the tangent calculated at the given point by finding dy/dx using parametric differentiation.
65 Easy
65 of 164
Application +4 | -1
If the tangent to the curve, y=f(x)=xlog⁡e  x,y = f\left( x \right) = x{\log _e}\,\,x, (x>0)(x > 0) at a point (c, f(c))(c,\, f(c)) is parallel to the line segment joining the point (1, 0) and (e,e)(e, e) then cc is equal to
View answer and explanation
Correct option: B
e(1e−1){e^{\left( {\frac{1}{{e - 1}}} \right)}}
Option B (e(1e−1){e^{\left( {\frac{{1}}{{e - 1}}} \right)}}) is correct because it is the value of cc that satisfies the condition of the tangent being parallel to the line segment.
66 Easy
66 of 164
Application +4 | -1
The tangent to the curve y=x3y=x^3 at the point P(t,t3)P\left(t, t^3\right) cuts the curve again at point QQ. Then, the coordinates of QQ are
View answer and explanation
Correct option: D
(−2t,−8t3)\left(-2 t,-8 t^3\right)
Option D ((-2t, -8t3)) is correct because it represents the other intersection point of the tangent line with the curve y=x3.
67 Easy
67 of 164
Application +4 | -1
The normal to the curve 5x5−10x3+x+2y+6=0 at P(0,  −3)5{x^5} - 10{x^3} + x + 2y + 6 = 0\,at\,P\left( {0,\; - 3} \right) meets the curve again at the point
View answer and explanation
Correct option: B
(1,  −1),  (−1,  −5)\left( {1,\; - 1} \right),\;\left( { - 1,\; - 5} \right)
Option D ((−1,  5),  (1,  −1)\left( { - 1,\;5} \right),\;\left( {1,\; - 1} \right)) is correct because by finding the equation of the normal and solving it simultaneously with the curve's equation, we find the intersection points.
68 Easy
68 of 164
Application +4 | -1
If the curves y2=6x, 9x2+by2=16{y^2} = 6x,\,9{x^2} + b{y^2} = 16 intersect each other at right angles, then the value of bb is
View answer and explanation
Correct option: D
92\frac{9}{2}
Option D (92\frac{9}{2}) is correct because the value of b is found by using the condition that the curves intersect at right angles, implying the product of their slopes at the point of intersection is -1.
69 Easy
69 of 164
Analyzing +4 | -1
The equation of the normal to the curve y=x−xy=x^{-x} at the point of its maximum is
View answer and explanation
Correct option: B
x=e−1x=e^{-1}
Option B (x=e−1x=e^{-1}) is correct because the normal to the curve at its maximum point is a vertical line at x=1/e.
70 Easy
70 of 164
Application +4 | -1
If the curve y=x2+bx+cy=x^2+b x+c touches the line y=xy=x at the point (1,1)(1,1), then the set of values of xx for which the curve has a negative gradient is
View answer and explanation
Correct option: A
(−∞,1/2)(-\infty, 1 / 2)
Option A ((−∞,1/2))((-\infty, 1 / 2)) is correct because finding the derivative of the curve and setting it less than 0 gives the interval where the gradient is negative.
71 Easy
71 of 164
Analyzing +4 | -1
If θ\theta is the angle between the curves xy=2xy = 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 is the absolute value of the tangent of the angle between the curves, calculated using their derivatives at the point of intersection.
72 Easy
72 of 164
Application +4 | -1
The tangent to the curve y=e2xy=e^{2 x} at the point (0,1)(0,1) meets xx-axis at
View answer and explanation
Correct option: C
(−1/2,0)(-1 / 2,0)
Option C (-1/2,0) is correct because it is the x-intercept of the tangent line to the curve at the point (0,1).
73 Easy
73 of 164
Analyzing +4 | -1
PP is a variable point on the curve y=f(x)y=f(x) and AA is a fixed point in the plane not lying on the curve. If PA2P A^2 is minimum, then the angle between PAP A and the tangent at PP is
View answer and explanation
Correct option: C
π2\frac{\pi}{2}
Option C (π2\frac{\pi}{2}) is correct because the minimum distance between a point and a curve is always along the normal to the curve.
74 Easy
74 of 164
Analyzing +4 | -1
The angle at which the curves y=sin⁡xy=\sin x and y=cos⁡xy=\cos x intersect in [0,π][0, \pi], is
View answer and explanation
Correct option: B
±tan⁡−122\pm \tan ^{-1} 2 \sqrt{2}
Option A (±tan⁡−122\pm \tan^{-1} 2 \sqrt{2}) is correct because it is the angle between the curves calculated using the derivatives at their intersection point.
75 Easy
75 of 164
Application +4 | -1
Tangent at point P(t,t3)P\left(t, t^3\right) on the curve y=x3y=x^3 meets the curve again at point QQ. If a point on PQP Q divides it internally in the ratio 1:21: 2, then the coordinates of that point are
View answer and explanation
Correct option: A
(0,−2t3)\left(0,-2 t^3\right)
Option A (0,−2t3)\left(0,-2 t^3\right) is correct because it represents the coordinates of the point that divides the line segment PQ in the ratio 1:2 internally.

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76 Easy
76 of 164
Application +4 | -1
If 2xy+3yx=202 x^{y}+3 y^{x}=20, then dydx\frac{d y}{d x} at (2,2)(2,2) is equal to:
View answer and explanation
Correct option: A
−(2+log⁡e83+log⁡e4)-\left(\frac{2+\log _{e} 8}{3+\log _{e} 4}\right)
Option A (−(2+log⁡e83+log⁡e4)-(\frac{{2+\log _{e} 8}}{{3+\log _{e} 4}})) is correct because it is the result of differentiating the given equation implicitly and then substituting the given point (2,2).
77 Easy
77 of 164
Application +4 | -1
The equation of the normal to the curve y=x+sin⁡xcos⁡xy=x+\sin x \cos x at x=π2x=\frac{\pi}{2}, is
View answer and explanation
Correct option: D
2x=π2 x=\pi
Option D (2x=π2x=\pi) is correct because the normal to the curve at the given point is a vertical line passing through x = pi/2.
78 Easy
78 of 164
Application +4 | -1
If the tangent at (1,1)(1,1) on y2=x(2−x)2y^2=x(2-x)^2 meets the curve again at PP, then PP is
View answer and explanation
Correct option: C
(9/4,3/8)(9 / 4,3 / 8)
Option C ((9/4,3/8)(9 / 4,3 / 8)) is correct because the tangent line equation is found, and then solved simultaneously with the original curve to find the intersection point P.
79 Easy
79 of 164
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.
80 Easy
80 of 164
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.
81 Easy
81 of 164
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.
82 Easy
82 of 164
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.
83 Easy
83 of 164
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).
84 Easy
84 of 164
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.
85 Easy
85 of 164
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.
86 Easy
86 of 164
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.
87 Easy
87 of 164
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.
88 Easy
88 of 164
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.
89 Easy
89 of 164
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.
90 Easy
90 of 164
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.
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