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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Mixed difficulty

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Showing 61-90 of 281 questions

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61 Easy
61 of 281
Application +4 | -1
At what point on the curve x3−8a2y=0{x^3} - 8{a^2}y = 0, the slope of the normal is −23- \frac{2}{3}?
View answer and explanation
Correct option: C
(2a,  a)\left( {2a,\;a} \right)
Option D (xa+yb=1\frac{x}{a} + \frac{y}{b} = 1) is correct because it represents the equation of the tangent to the curve at the point where it crosses the y-axis.
62 Easy
62 of 281
Application +4 | -1
If the tangent to the curve 2y3=ax2+x32 y^3=a x^2+x^3 at the point (a,a)(a, a) cuts off intercepts α\alpha and β\beta on the coordinate axes such that α2+β2=61\alpha^2+\beta^2=61, then a=a=
View answer and explanation
Correct option: A
±30\pm 30
Option A (±30\pm 30) is correct because substituting the given conditions into the equation and its derivative allows us to solve for a.
63 Easy
63 of 281
Application +4 | -1
For the curve x=3cos⁡θ,y=3sin⁡θx=3 \cos \theta, y=3 \sin \theta, 0≤θ≤π,0 \leq \theta \leq \pi, the tangent is parallel to the xx-axis, where θ=\theta=
View answer and explanation
Correct option: D
π2\frac{\pi}{2}
Option D (π2\frac{\pi}{2}) is correct because at this angle, the derivative dydx\frac{dy}{dx} equals zero, indicating a tangent parallel to the x-axis.
64 Easy
64 of 281
Analyzing +4 | -1
The angle between one-sided tangents to the curve y=e∣x∣y=e^{|x|} at x=0x=0, is
View answer and explanation
Correct option: C
π2\frac{\pi}{2}
Option C (π2\frac{{\pi}}{{2}}) is correct because the slopes of the one-sided tangents are -1 and 1, which are negative reciprocals, indicating a right angle.
65 Easy
65 of 281
Application +4 | -1
The normal at point (1,1) of the curve y2=x3{y^2} = {x^3} is parallel to the line
View answer and explanation
Correct option: B
2x+3y−7=02x + 3y - 7 = 0
Option B (2x+3y−7=02x + 3y - 7 = 0) is correct because the normal to the curve at (1,1) is parallel to this line.
66 Easy
66 of 281
Application +4 | -1
The slope of the normal at the point with abscissa x=−2x=-2 of the graph of the function f(x)=∣x2−x∣f(x)=\left|x^{2}-x\right| is
View answer and explanation
Correct option: D
1/31 / 3
Option D (1/3) is correct because it is the negative reciprocal of the derivative of the function evaluated at x = -2, considering the absolute value.
67 Easy
67 of 281
Application +4 | -1
The sum of the intercepts made on the axes of coordinates by any tangent to the curve x+y=a\sqrt{x}+\sqrt{y}=\sqrt{a} is equal to
View answer and explanation
Correct option: B
aa
Option B (aa) is correct because the sum of the intercepts is equal to a(x1+y1)\sqrt{a}(\sqrt{x_1} + \sqrt{y_1}), which simplifies to aa using the equation of the curve.
68 Easy
68 of 281
Application +4 | -1
A function f(x)f(x) is such that f′′(x)=6(x−1)f^{\prime \prime}(x)=6(x-1). If the graph passes through the point (2,1)(2,1) and at that point the tangent is y=3x−5y=3 x-5, then f(x)=f(x)=
View answer and explanation
Correct option: B
(x−1)3(x-1)^3
Option B ((x−1)3(x-1)^3) is correct because the function satisfies the given conditions: its second derivative is 6(x-1), it passes through (2,1), and its tangent at that point is y=3x-5.
69 Easy
69 of 281
Application +4 | -1
If the tangent to the curve xy+ax+by=0x y+a x+b y=0 at (1,1)(1,1) is inclined at an angle tan⁡−12\tan ^{-1} 2 with xx-axis, then
View answer and explanation
Correct option: B
a=1,b=−2a=1, b=-2
The correct answer is A (a=1,b=−2a=1, b=-2) because these values satisfy both the condition that (1,1) lies on the curve and the condition that the tangent at (1,1) has a slope of 2.
70 Easy
70 of 281
Analyzing +4 | -1
The two tangents to the curve ax2+2hxy+by2=1,  a>0a{x^2} + 2hxy + b{y^2} = 1,\;a > 0 at the points where it crosses xx-axis, are
View answer and explanation
Correct option: A
Parallel
Option B (a∈(−6,;6)a \in ( - \sqrt 6 ,\\;\sqrt 6 )) is correct because it ensures that the function is decreasing for all x, which requires the derivative to be negative.
71 Medium
71 of 281
Application +4 | -1
If (a,b)(a, b) be the point on the curve 9y2=x39 y^{2}=x^{3} where normal to the curve make equal intercepts with the axis, then the value of (a+b)(a+b) is :
View answer and explanation
Correct option: C
203\frac{20}{3}
Option C (203\frac{20}{3}) is correct because finding the derivative, setting the normal's slope to -1, and solving the resulting equations yields the correct a and b values.
72 Easy
72 of 281
Application +4 | -1
The equation of the tangents to 2x2−3y2=362{x^2} - 3{y^2} = 36 which are parallel to the straight line x+2y−10=0x + 2y - 10 = 0 are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculation leads to imaginary values for y, indicating that no real tangents exist that satisfy the given conditions.
73 Easy
73 of 281
Understanding +4 | -1
For a curve  (length of normal)  (length of tangent) \frac{\text { (length of normal) }}{\text { (length of tangent) }} is equal to
View answer and explanation
Correct option: C
Slope of tangent
Option C (Slope of tangent) is correct because the ratio of the length of the normal to the length of the tangent is equal to the slope of the tangent.
74 Easy
74 of 281
Application +4 | -1
If the equation of the tangent to the curve y2=ax3+by^2=a x^3+b at the point (2,3)(2,3) is y=4x−5y=4 x-5, then
View answer and explanation
Correct option: C
a=2,b=−7a=2, b=-7
Option C (a=2,b=−7a=2, b=-7) is correct because substituting the point (2,3) and the tangent equation into the curve equation and its derivative yields these values for a and b.
75 Easy
75 of 281
Understanding +4 | -1
The tangent to a given curve is perpendicular to xx-axis, if
View answer and explanation
Correct option: C
dxdy=0\frac{d x}{d y}=0
Option C (dxdy=0\frac{d x}{d y}=0) is correct because if the tangent is perpendicular to the x-axis, it means the tangent is vertical, and the slope dydx\frac{dy}{dx} is undefined, which implies dxdy=0\frac{dx}{dy} = 0.

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76 Medium
76 of 281
Application +4 | -1
The curve xnan+ynbn=2\frac{x^n}{a^n}+\frac{y^n}{b^n}=2 touches the line xa+yb=2\frac{x}{a}+\frac{y}{b}=2 at the point
View answer and explanation
Correct option: B
(a,b)(a, b)
Option B ((a,b)(a, b)) is correct because the curve touches the line at the point (a,b)(a, b), which can be found by equating the slopes and using the equation of the curve.
77 Hard
77 of 281
Application +4 | -1
For the curve y=3sin⁡θcos⁡θ,x=eθsin⁡θ\mathrm{y}=3 \sin \theta \cos \theta, \mathrm{x}=\mathrm{e}^\theta \sin \theta, 0≤θ≤π0 \leq \theta \leq \pi, the tangent is parallel to xx-axis when θ\theta is:
View answer and explanation
Correct option: C
π4\frac{\pi}{4}
Option C (π4\frac{{\pi}}{{4}}) is correct because the tangent is parallel to the x-axis when dydx=0\frac{{dy}}{{dx}} = 0, which occurs when θ=π4\theta = \frac{{\pi}}{{4}} after differentiating and simplifying.
78 Easy
78 of 281
Application +4 | -1
The equation of the normal to the curve y=x(2−x)y=x(2-x) at the point (2,0)(2,0) is
View answer and explanation
Correct option: A
x−2y=2x-2 y=2
Option A (x−2y=2x-2y=2) is correct because it represents the equation of the normal line to the curve at the point (2,0).
79 Easy
79 of 281
Application +4 | -1
If the tangent at a point PP, with parameter tt, on the curve x=4t2+3,y=8t3−1,t∈Rx=4 t^2+3, y=8 t^3-1, t \in R, meets the curve again at QQ, then the coordinates of QQ are
View answer and explanation
Correct option: A
(t2+3,−t3−1)\left(t^2+3,-t^3-1\right)
Option A (t2+3,−t3−1)\left(t^2+3,-t^3-1\right) is correct because it is found by determining the intersection point of the tangent line at parameter t with the original curve.
80 Easy
80 of 281
Application +4 | -1
The co-ordinates of the point on the curve y=x2−3x+2y=x^2-3 x+2 where the tangent is perpendicular to the straight line y=xy=x are
View answer and explanation
Correct option: B
(1,0)(1,0)
Option B (1,0) is correct because it is the point on the curve where the tangent's slope is -1, which is perpendicular to the line y=x.
81 Easy
81 of 281
Analyzing +4 | -1
The function f(x)=xxf(x)=x^x has a stationary point at
View answer and explanation
Correct option: B
x=1ex=\frac{1}{e}
Option B (x=1ex=\frac{1}{e}) is correct because it is the value of x where the derivative of the function f(x)=xxf(x) = x^x is equal to zero, indicating a stationary point.
82 Easy
82 of 281
Application +4 | -1
If λx+my=1\lambda x+m y=1 touches the curve (ax)n+( by )n=1(a x)^n+(\text { by })^n=1. If (ℓa)nn−1+(mb)nn−1=p\left(\frac{\ell}{a}\right)^{\frac{n}{n-1}}+\left(\frac{m}{b}\right)^{\frac{n}{n-1}}=p, then value of pp is
View answer and explanation
Correct option: A
1
Option A (1) is correct because by comparing the given tangent equation with the general tangent equation of the curve, the value of p is found to be 1.
83 Easy
83 of 281
Analyzing +4 | -1
Let f(x)=sin⁡x−tan⁡x,x∈(0,π/2)f(x)=\sin x-\tan x, x \in(0, \pi / 2) then tangent drawn to the curve y=f(x)y=f(x) at any point will
View answer and explanation
Correct option: A
Lie above the curve
Option A (Lie above the curve) is correct because the second derivative is negative, indicating the curve is concave down, so the tangent lies above.
84 Easy
84 of 281
Application +4 | -1
If the tangent to the curve xy+ax+by=0x y+a x+b y=0 at (1,1)(1,1) makes an angle of measure tan⁡−12\tan ^{-1} 2 with positive X-axis, then a+bab=\frac{a+b}{a b}=
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because by differentiating the given equation and using the given information, we can solve for a and b and then calculate the required expression.
85 Medium
85 of 281
Application +4 | -1
The distance, from the origin, of the normal to the curve, x=2cos⁡t+2tsin⁡t,y=2sin⁡t−2tcos⁡tx=2 \cos t+2 t \sin t, y=2 \sin t-2 t \cos t at t=π4t=\frac{\pi}{4}, is :
View answer and explanation
Correct option: A
2
Option A (2) is correct because the distance from the origin to the normal is calculated correctly using the given parametric equations and the formula for the distance from a point to a line.
86 Medium
86 of 281
Application +4 | -1
A tangent is drawn at a variable point on an ellipse x2/a2+y2/b2=1x^2 / a^2+y^2 / b^2=1, then minimum area of the triangle formed by the tangent and the coordinate axes is
View answer and explanation
Correct option: A
aba b
Option A (abab) is correct because the minimum area of the triangle formed by the tangent and the coordinate axes is abab.
87 Medium
87 of 281
Application +4 | -1
The normal to the curve y(x−2)(x−3)=x+6y\left( {x - 2} \right)\left( {x - 3} \right) = 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 ((12,12)\left( {\frac{1}{2},\frac{1}{2}} \right)) is correct because the normal to the curve at the y-intercept passes through this point.
88 Easy
88 of 281
Application +4 | -1
The slope of the tangent to the curve (y−x5)2=x(1+x2)2\left(y-x^5\right)^2=x\left(1+x^2\right)^2 at the point (1,3)(1,3), is
View answer and explanation
Correct option: C
8
Option C (8) is correct because differentiating the curve equation and substituting the point (1,3) yields a slope of 8.
89 Easy
89 of 281
Application +4 | -1
For which of the following curves, the line x+3y=23x + \sqrt 3 y = 2\sqrt 3 is the tangent at the point (332,    12)\left( {\frac{{3\sqrt 3 }}{2},\,\,\,\,\frac{1}{2}} \right) ?
View answer and explanation
Correct option: D
x2+9y2=9{x^2} + 9{y^2} = 9
Option D (x2+9y2=9{x^2} + 9{y^2} = 9) is the correct curve because the given line is tangent to it at the specified point.
90 Medium
90 of 281
Application +4 | -1
The chord joining the points where x=px = p and x=qx = q on the curve y=ax2+bx+cy = a{x^2} + bx + c is parallel to the tangent at the point on the curve whose abscissa is
View answer and explanation
Correct option: A
p+q2\frac{{p + q}}{2}
Option A (p+q2\frac{{p + q}}{2}) is correct because it represents the abscissa where the tangent is parallel to the chord joining the points on the curve, found by equating the slope of the chord to the derivative of the curve.
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