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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Mixed difficulty

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Showing 61-90 of 429 questions

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61 Medium
61 of 429
Application +4 | -1
Foci of the hyperbola x216−(y−2)29=1\frac{x^2}{16}-\frac{(y-2)^2}{9}=1 are
View answer and explanation
Correct option: A
(5,2)(−5,2)(5,2)(-5,2)
Option A ((5,25,2), (−5,2)(-5,2)) is correct because these are the foci calculated using the eccentricity and center of the given hyperbola.
62 Easy
62 of 429
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.
63 Easy
63 of 429
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.
64 Medium
64 of 429
Application +4 | -1
If the line y=mx+73y = mx + 7\sqrt 3 is normal to the hyperbola x224−y218=1,\frac{{{x^2}}}{{24}} - \frac{{{y^2}}}{{18}} = 1, then a value of mm is
View answer and explanation
Correct option: C
25\frac{2}{{\sqrt 5 }}
Option C (25\frac{2}{{\sqrt 5 }}) is correct because it is a value of m for which the line y=mx+73y = mx + 7\sqrt 3 is normal to the hyperbola x224−y218=1\frac{{{x^2}}}{{24}} - \frac{{{y^2}}}{{18}} = 1.
65 Easy
65 of 429
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.
66 Easy
66 of 429
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.
67 Easy
67 of 429
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.
68 Easy
68 of 429
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.
69 Easy
69 of 429
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.
70 Easy
70 of 429
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.
71 Easy
71 of 429
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.
72 Easy
72 of 429
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}.
73 Easy
73 of 429
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.
74 Easy
74 of 429
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.
75 Easy
75 of 429
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.

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76 Easy
76 of 429
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.
77 Easy
77 of 429
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.
78 Medium
78 of 429
Analyzing +4 | -1
A hyperbola touches y – axis and has its centre and one of the foci are at (520,20)\left( {\frac{5}{{20}},20} \right) and
(10, 24) respectively; then length of its transverse axis is
View answer and explanation
Correct option: C
89\sqrt {89}
Option C (89\sqrt {89}) is the correct answer because it is derived from the geometric properties of the hyperbola and the given conditions.
79 Hard
79 of 429
Analyzing +4 | -1
The circle x2+y2−8x=0x^2+y^2-8 x=0 and hyperbola x29−y24=1\frac{x^2}{9}-\frac{y^2}{4}=1 intersect at the points AA and BB. The equation of the circle with ABA B as its diameter, is
View answer and explanation
Correct option: A
x2+y2−12x+24=0x^2+y^2-12 x+24=0
Option A (x2+y2−12x+24=0x^2+y^2-12 x+24=0) is correct because it represents the equation of the circle with AB as its diameter, where A and B are the intersection points of the given circle and hyperbola.
80 Easy
80 of 429
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.
81 Easy
81 of 429
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.
82 Medium
82 of 429
Application +4 | -1
Tangent at point PP of hyperbola xy=14x y=14 intersects the coordinate axes at AA and BB. The locus of centroid of △OAB\triangle \mathrm{OAB} where OO is origin, is
View answer and explanation
Correct option: D
a hyperbola
Option D (a hyperbola) is correct because the locus of the centroid of the triangle formed by the tangent to the hyperbola and the coordinate axes is a hyperbola.
83 Medium
83 of 429
Application +4 | -1
The equation of the chord of contact of tangents drawn from a point (2,−1)(2,-1) to the hyperbola 16x2−9y2=14416 x^2-9 y^2=144, is
View answer and explanation
Correct option: C
32x+9y=14432 x+9 y=144
Option C (32x+9y=14432 x+9 y=144) is correct because it is the equation of the chord of contact of tangents drawn from the given point to the hyperbola.
84 Easy
84 of 429
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.
85 Easy
85 of 429
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.
86 Easy
86 of 429
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.
87 Hard
87 of 429
Analyzing +4 | -1
If the chords of the rectangular hyperbola x2−y2=a2{x^2} - {y^2} = {a^2} touch the parabola y2=4ax,{y^2} = 4ax, then the locus of their mid-points is
View answer and explanation
Correct option: B
y2(x−a)=x3{y^2}\left( {x - a} \right) = {x^3}
Option B (y2(x−a)=x3{y^2}\left( {x - a} \right) = {x^3}) is correct because it represents the locus of the mid-points of the chords of the rectangular hyperbola that touch the given parabola.
88 Easy
88 of 429
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.
89 Easy
89 of 429
Application +4 | -1
Tangents are drawn to the hyperbola 5x2−y2=55 x^2-y^2=5 from the point (0,2)(0,2). Their equations are
View answer and explanation
Correct option: A
y=±3x+2y= \pm 3 x+2
Option A (y=±3x+2y= \pm 3 x+2) is correct because substituting the point (0,2) into the general tangent equation of the hyperbola yields the given equations.
90 Easy
90 of 429
Analyzing +4 | -1
Let P(6,3)P(6,3) be a point on the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. If the normal at the point PP intersects the xx-axis at (9,0)(9,0), then the eccentricity of the hyperbola, is
View answer and explanation
Correct option: B
32\sqrt{\frac{3}{2}}
Option B (32\sqrt{\frac{3}{2}}) is correct because by using the equation of the normal to the hyperbola and the given point, we can solve for the eccentricity.
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