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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Easy

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Showing 31-60 of 298 questions

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31 Easy
31 of 298
Application +4 | -1
Equation of the normal to the hyperbola x225−y216=1\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1 perpendicular to the line 2x+y=12x + y = 1 is
View answer and explanation
Correct option: A
21(x−2y)=41\sqrt {21} \left( {x - 2y} \right) = 41
Option A (21(x−2y)=41\sqrt {21} \left( {x - 2y} \right) = 41) is correct because it represents the equation of the normal to the hyperbola that is perpendicular to the given line, found by using trigonometric substitution and the condition for perpendicularity.
32 Easy
32 of 298
Application +4 | -1
The distance between the foci of the hyperbola x2−3y2−4x−6y−11=0{x^2} - 3{y^2} - 4x - 6y - 11 = 0 is
View answer and explanation
Correct option: C
8
Option C (8) is correct because after rewriting the equation in standard form, the distance between the foci is calculated as 2ae, which equals 8.
33 Easy
33 of 298
Application +4 | -1
The equation of the hyperbola with vertices (3,0)(3,0) and (−3,0)(-3,0) and semi-latusrectum 4, is given by
View answer and explanation
Correct option: C
4x2−3y2−36=04 x^2-3 y^2-36=0
Option C (4x2−3y2−36=04 x^2-3 y^2-36=0) is correct because it represents a hyperbola with vertices at (±3,0)(\pm 3, 0) and the given semi-latus rectum.
34 Easy
34 of 298
Application +4 | -1
The equation of the tangent to the hyperbola 2x2−3y2=62 x^2-3 y^2=6 which is parallel to the line y=3x+4y=3 x+4, is
View answer and explanation
Correct option: C
y=3x+5y=3 x+5 and y=3x−5y=3 x-5
Option C (y=3x+5y=3x+5 and y=3x−5y=3x-5) is correct because by finding the slope of the given line and using the tangent condition for a hyperbola, we can determine the equations of the tangents parallel to the line.
35 Easy
35 of 298
Application +4 | -1
The length of transverse axis of the hyperbola 3x2−4y2=323{x^2} - 4{y^2} = 32 is
View answer and explanation
Correct option: A
823\frac{{8\sqrt 2 }}{{\sqrt 3 }}
Option A (823\frac{{8\sqrt 2 }}{{\sqrt 3 }}) is correct because it represents the length of the transverse axis of the given hyperbola after rewriting the equation in standard form.
36 Easy
36 of 298
Application +4 | -1
The angle between the asymptotes of the hyperbola x216−y29=1,\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1, is
View answer and explanation
Correct option: A
π−2tan⁡−134\pi - 2{\tan ^{ - 1}}\frac{3}{4}
Option A (π−2tan⁡−134\pi - 2{\tan ^{ - 1}}\frac{3}{4}) is correct because it is derived from the formula for the angle between the asymptotes of a hyperbola.
37 Easy
37 of 298
Analyzing +4 | -1
The tangents from a point (22,1)\left( {2\sqrt 2 ,1} \right) to the hyperbola 16x2−25y2=40016{x^2} - 25{y^2} = 400 include an angle equal to
View answer and explanation
Correct option: A
π2\frac{\pi }{2}
Option A (π2\frac{\pi }{2}) is correct because the point lies on the director circle of the hyperbola, implying the tangents are perpendicular.
38 Easy
38 of 298
Understanding +4 | -1
If the length of the transverse and conjugate axes of a hyperbola be 8 and 6 , respectively, then the difference in the focal distances of any point of the hyperbola is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the difference in the focal distances of any point on a hyperbola is equal to the length of the transverse axis, which is given as 8.
39 Easy
39 of 298
Application +4 | -1
The directrix of the hyperbola x29−y24=1\frac{{{x^2}}}{9} - \frac{{{y^2}}}{4} = 1 is
View answer and explanation
Correct option: D
x=913x = \frac{9}{{\sqrt {13} }}
Option D (x=913x = \frac{9}{{\sqrt {13} }}) is correct because the directrix of the hyperbola is calculated using the formula x=±aex = \pm \frac{a}{e}.
40 Easy
40 of 298
Application +4 | -1
If asymptotes of a hyperbola are at 900{90^0}, then
View answer and explanation
Correct option: A
Eccentricity is 2\sqrt 2
Option A (Eccentricity is 2\sqrt 2) is correct because a hyperbola with asymptotes at 90 degrees is a rectangular hyperbola, and its eccentricity is always 2\sqrt 2.
41 Easy
41 of 298
Application +4 | -1
The equation of the chord of contact of tangents from (1,  2)\left( {1,\;2} \right) to the hyperbola 3x2−4y2=33{x^2} - 4{y^2} = 3 is
View answer and explanation
Correct option: B
3x−8y−3=03x - 8y - 3 = 0
Option B (3x−8y−3=03x - 8y - 3 = 0) is correct because the equation of the chord of contact of tangents from a point to a hyperbola is given by T=0T = 0.
42 Easy
42 of 298
Application +4 | -1
If the line y=mx+a2m2−b2y=m x+\sqrt{a^2 m^2-b^2} touches the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 at the point φ\varphi. Then φ=\varphi=
View answer and explanation
Correct option: C
sin⁡−1(bam)\sin ^{-1}\left(\frac{{b}}{{am}}\right)
Option C (sin⁡−1(bam)\sin ^{-1}\left(\frac{{b}}{{am}}\right)) is correct because it represents the point of tangency on the hyperbola.
43 Easy
43 of 298
Application +4 | -1
The equation of the hyperbola referred to the axes of coordinate and whose distance between the foci is 16 and eccentricity is 2\sqrt 2, is
View answer and explanation
Correct option: B
x2−y2=32{x^2} - {y^2} = 32
Option B (x2−y2=32{x^2} - {y^2} = 32) is correct because it is derived from the given distance between the foci and the eccentricity of the hyperbola.
44 Easy
44 of 298
Application +4 | -1
If the line y=mx+cy=m x+c is a common tangent to the hyperbola x2100−y264=1\frac{x^2}{100}-\frac{y^2}{64}=1 and the circle x2+y2=36x^2+y^2=36, then which one of the following is true?
View answer and explanation
Correct option: A
4c2=3694 c^2=369
Option A (4c2=3694 c^2=369) is correct because it correctly uses the tangency conditions for both the hyperbola and the circle to find the relationship between c and m.
45 Easy
45 of 298
Application +4 | -1
If the line lx+my=1lx + my = 1 is a normal to the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1, then a2l2−b2m2\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} is equal to
View answer and explanation
Correct option: C
(a2+b2)2{\left( {{a^2} + {b^2}} \right)^2}
Option C ((a2+b2)2{\left( {{a^2} + {b^2}} \right)^2}) is correct because it is derived from the condition for a line to be a normal to the hyperbola.

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46 Easy
46 of 298
Analyzing +4 | -1
The locus of the mid – points of the chords of the hyperbola x2−y2=4{x^2} - {y^2} = 4, which touch the parabola y2=8x{y^2} = 8x, is
View answer and explanation
Correct option: C
y2(x−2)=x3{y^2}\left( {x - 2} \right) = {x^3}
Option C (y2(x−2)=x3{y^2}\left( {x - 2} \right) = {x^3}) is correct because it represents the locus of the mid-points of the chords of the hyperbola that touch the parabola.
47 Easy
47 of 298
Application +4 | -1
The coordinates of the point of intersection of tangents drawn to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 at the points where it is intersected by the line lx+my+n=0l x+m y+n=0 are
View answer and explanation
Correct option: B
(−a2ln,b2mn)\left(-\frac{a^2 l}{n}, \frac{b^2 m}{n}\right)
Option B ((−a2ln,b2mn)\left(-\frac{a^2 l}{n}, \frac{b^2 m}{n}\right)) is correct because it represents the coordinates of the intersection point of tangents, derived by comparing the chord of contact equation with the given line equation.
48 Easy
48 of 298
Application +4 | -1
If the distance between two directrices of a rectangular hyperbola is 15, then the distance between its foci in units is:
View answer and explanation
Correct option: B
30
Option B (30) is correct because using the rectangular hyperbola's eccentricity and the directrices distance, the foci distance is calculated as 30.
49 Easy
49 of 298
Application +4 | -1
The angle between the asymptotes of the hyperbola 3x2−y2=33{x^2} - {y^2} = 3 is
View answer and explanation
Correct option: C
2π3\frac{{2\pi }}{3}
Option C (2π3\frac{{2\pi }}{{3}}) is correct because the angle between the asymptotes is 2tan⁡−1(ba)2\tan^{{-1}}(\frac{b}{a}), where a=1a=1 and b=3b=\sqrt{3}.
50 Easy
50 of 298
Application +4 | -1
The eccentricity of the conjugate hyperbola of the hyperbola x2−3y2=1x^2-3 y^2=1, is
View answer and explanation
Correct option: A
2
The correct answer is A (2) because the eccentricity of the conjugate hyperbola is calculated using the relationship between the eccentricities of the hyperbola and its conjugate.
51 Easy
51 of 298
Application +4 | -1
The eccentricity of the hyperbola whose asymptotes are 3x+4y=23x + 4y = 2 and 4x−3y+5=04x - 3y + 5 = 0, is
View answer and explanation
Correct option: C
2\sqrt 2
Option C (2\sqrt 2) is correct because the asymptotes are perpendicular, indicating a rectangular hyperbola, which always has an eccentricity of 2\sqrt 2.
52 Easy
52 of 298
Application +4 | -1
The equation of the tangent to the conic x2−y2−8x+2yx^2-y^2-8 x+2 y +11=0+11=0 at the point (2,1)(2,1) is
View answer and explanation
Correct option: C
x−2=0x-2=0
Option C (x−2=0x-2=0) is correct because substituting the point (2,1) into the tangent equation yields the correct result, x=2.
53 Easy
53 of 298
Application +4 | -1
The equation x2−16xy−11y2−12x+6y+21=0x^2-16 x y-11 y^2-12 x+6 y+21=0 represents
View answer and explanation
Correct option: C
Hyperbola
Option C (Hyperbola) is correct because the condition h2>abh^2 > ab is satisfied, indicating a hyperbola.
54 Easy
54 of 298
Application +4 | -1
If the eccentricity of a hyperbola x29−y2b2=1\frac{x^{2}}{9}-\frac{y^{2}}{b^{2}}=1, which passes through (k,2)(k, 2), is 133\frac{\sqrt{13}}{3}, then the value of k2k^{2} is
View answer and explanation
Correct option: A
18
Option A (18) is correct because the value of k2k^2 is calculated using the eccentricity and the point through which the hyperbola passes.
55 Easy
55 of 298
Application +4 | -1
The line 3x−4y=53 x-4 y=5 is a tangent to the hyperbola x2−4y2=5x^2-4 y^2=5. The point of contact is
View answer and explanation
Correct option: A
(3,1)(3,1)
Option A ((3,1)) is correct because it satisfies both the hyperbola equation and the tangent equation.
56 Easy
56 of 298
Application +4 | -1
The line xcos⁡α+ysin⁡α=px \cos \alpha+y \sin \alpha=p will be a tangent to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 if
View answer and explanation
Correct option: B
a2cos⁡2α−b2sin⁡2α=p2a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2
Option B (a2cos⁡2α−b2sin⁡2α=p2a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2) is correct because this is the condition for the line to be a tangent to the hyperbola.
57 Easy
57 of 298
Understanding +4 | -1
The number of points outside the hyperbola x29−y216=1\frac{x^2}{9}-\frac{y^2}{16}=1 from where two perpendicular tangents can be drawn to the hyperbola are
View answer and explanation
Correct option: A
0
Option A (0) is correct because the director circle of the hyperbola is imaginary, meaning there are no real points from which perpendicular tangents can be drawn.
58 Easy
58 of 298
Application +4 | -1
If (a−2)x2+ay2=4(a-2) x^2+a y^2=4 represents rectangular hyperbola, then a equals
View answer and explanation
Correct option: A
1
Option A (1) is correct because for a rectangular hyperbola, the coefficients of x2x^2 and y2y^2 must be equal in magnitude and opposite in sign, leading to the equation a−2=−aa-2 = -a, which solves to a=1a=1.
59 Easy
59 of 298
Application +4 | -1
If the equation of the hyperbola is 9y2−4x2=369 {y}^2-4 {x}^2=36, then

I. the coordinates of foci are (0,±13)(0, \pm \sqrt{13})

II. the eccentricity is 213\frac{2}{\sqrt{13}}

III. the length of the latus rectum is 8
View answer and explanation
Correct option: A
Only I is true
Option A (Only I is true) is correct because the coordinates of the foci are indeed (0, ±13\pm \sqrt{13}) as calculated from the hyperbola equation.
60 Easy
60 of 298
Application +4 | -1
The angle between the asymptotes of the hyperbola 2x2−2y2=92{x^2} - 2{y^2} = 9 is
View answer and explanation
Correct option: D
π2\frac{\pi }{2}
Option D (π2\frac{\pi }{2}) is correct because the given equation represents a rectangular hyperbola, and the angle between the asymptotes of a rectangular hyperbola is always 90 degrees.
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