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JEE • MATHEMATICS • TANGENTS & NORMALS

Tangents & normals practice · Easy

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Showing 31-60 of 164 questions

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31 Easy
31 of 164
Application +4 | -1
The abscissa of the point on the curve

y=a(ex/a+e−x/a)y = a\left( {{e^{x/a}} + {e^{ - x/a}}} \right)

Where the tangent is parallel to the x-axis, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the derivative of the given function, representing the slope of the tangent, is zero at x=0, indicating a horizontal tangent.
32 Easy
32 of 164
Application +4 | -1
The equation of the tangent to the curve y=4e−−x4  y = 4{e^{ - - \frac{x}{4}}}\; at the point where the curve crosses yy-axis is equal to
View answer and explanation
Correct option: C
x+y=4x + y = 4
Option C (x+y=4x + y = 4) is correct because it represents the equation of the tangent line to the curve at the point where it crosses the y-axis, calculated using the derivative.
33 Easy
33 of 164
Application +4 | -1
If x+4y=14x+4 y=14 is a normal to the curve y2=αx3−βy^{2}=\alpha x^{3}-\beta at (2,3)(2,3), then value of α+β\alpha+\beta is :
View answer and explanation
Correct option: A
9
Option A (9) is correct because substituting the point (2,3) into the curve equation and using the slope of the normal line allows us to solve for alpha and beta, resulting in alpha + beta = 9.
34 Easy
34 of 164
Analyzing +4 | -1
The curve x=y2 and xy=kx = {y^2}\,\mathrm{and}\,xy = k cut at right angles, if
View answer and explanation
Correct option: D
8k2=18{k^2} = 1
Option D (8k2=18k^2 = 1) is correct because it satisfies the condition for orthogonal intersection of the two curves, derived by equating the product of their slopes to -1.
35 Easy
35 of 164
Application +4 | -1
A normal to the hyperbola, 4x2−9y2=364 x^2-9 y^2=36 meets the coordinate axes xx and yy at AA and BB, respectively. If the parallelogram OABP(OO A B P(O being the origin) is formed, then the locus of PP is
View answer and explanation
Correct option: C
9x2−4y2=1699 x^2-4 y^2=169
Option C (9x2−4y2=1699 x^2-4 y^2=169) is correct because it represents the locus of point P derived from the properties of the normal to the hyperbola and the parallelogram OABP.
36 Easy
36 of 164
Analyzing +4 | -1
If OTOT and ONON are perpendiculars dropped from the origin to the tangent and normal to the curve x=asin⁡3tx=a \sin ^3 t, y=acos⁡3ty=a \cos ^3 t at an arbitrary point, then which one of the following is correct?
View answer and explanation
Correct option: D
All the above
Option D (All the above) is correct because all the given statements regarding the curve and its tangent/normal are true.
37 Easy
37 of 164
Application +4 | -1
For the curve x=t2−1,y=t2−tx=t^2-1, y=t^2-t, the tangent line is perpendicular to xx-axis when
View answer and explanation
Correct option: A
t=0t=0
The correct answer is t=0t=0 because at this value, the tangent line to the curve is perpendicular to the x-axis, meaning dxdy=0\frac{dx}{dy} = 0.
38 Easy
38 of 164
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.
39 Easy
39 of 164
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.
40 Easy
40 of 164
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.
41 Easy
41 of 164
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.
42 Easy
42 of 164
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.
43 Easy
43 of 164
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.
44 Easy
44 of 164
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.
45 Easy
45 of 164
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.

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46 Easy
46 of 164
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.
47 Easy
47 of 164
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.
48 Easy
48 of 164
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.
49 Easy
49 of 164
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.
50 Easy
50 of 164
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.
51 Easy
51 of 164
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.
52 Easy
52 of 164
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).
53 Easy
53 of 164
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.
54 Easy
54 of 164
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.
55 Easy
55 of 164
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.
56 Easy
56 of 164
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.
57 Easy
57 of 164
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.
58 Easy
58 of 164
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.
59 Easy
59 of 164
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.
60 Easy
60 of 164
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.
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