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JEE • MATHEMATICS • HYPERBOLA

Hyperbola practice · Mixed difficulty

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Showing 391-420 of 429 questions

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391 Easy
391 of 429
Understanding +4 | -1
The difference in focal distances of any point on the hyperbola x216−y29=1\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the difference in focal distances of any point on a hyperbola is equal to the length of its transverse axis (2a).
392 Easy
392 of 429
Understanding +4 | -1
A point on the curve x2A2−y2B2=1\frac{x^2}{A^2}-\frac{y^2}{B^2}=1 is
View answer and explanation
Correct option: B
(Asec⁡θ,Btan⁡θ)(A \sec \theta, B \tan \theta)
Option B ((Asec⁡θ,Btan⁡θ)(A \sec \theta, B \tan \theta)) is correct because substituting x=Asec⁡θx = A \sec \theta and y=Btan⁡θy = B \tan \theta into the equation x2A2−y2B2=1\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 satisfies the equation.
393 Easy
393 of 429
Application +4 | -1
The equation of the tangent to the hyperbola 4y2=x2−14{y^2} = {x^2} - 1 at the point (1,0), is
View answer and explanation
Correct option: A
x=1x = 1
Option A (x=1x = 1) is the correct answer because substituting the point (1,0) into the equation of the tangent yields x=1x = 1.
394 Easy
394 of 429
Application +4 | -1
If in a hyperbola, the distance between the foci is 10 and the transverse axis has length 8, then the length of its latusrectum is
View answer and explanation
Correct option: B
92\frac{9}{2}
Option B (92\frac{9}{2}) is correct because using the given information to find 'a' and 'b' and then applying the latus rectum formula results in 9/2.
395 Medium
395 of 429
Application +4 | -1
The area of the triangle formed by any tangent to the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 with its asymptotes is
View answer and explanation
Correct option: D
abab
Option D (abab) is correct because the area of the triangle formed by the tangent to the hyperbola and its asymptotes is constant and equal to abab.
396 Easy
396 of 429
Application +4 | -1
If the chords of contact of tangents from two points (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 are at right angles, then x1x2y1y2\frac{x_1 x_2}{y_1 y_2} is equal to
View answer and explanation
Correct option: D
−a4b4-\frac{a^4}{b^4}
Option D (−a4b4-\frac{a^4}{b^4}) is correct because the product of the slopes of the chords of contact is -1, leading to the given ratio.
397 Easy
397 of 429
Analyzing +4 | -1
The line 2x+y=12 x+y=1 is tangent to the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. If this line passes through the point of intersection of the nearest directrix and the xx-axis, then the eccentricity of the hyperbola is
View answer and explanation
Correct option: B
2
Option B (2) is correct because by using the tangent condition and the fact that the line passes through the directrix intersection with the x-axis, we can solve for the eccentricity.
398 Medium
398 of 429
Application +4 | -1
A common tangent to 9x2−16y2=1449{x^2} - 16{y^2} = 144 and x2+y2=9{x^2} + {y^2} = 9, is
View answer and explanation
Correct option: B
y=327x+157y = 3\sqrt {\frac{2}{7}} x + \frac{{15}}{{\sqrt 7 }}
Option B (y=327x+157y = 3\sqrt {\frac{2}{7}} x + \frac{{15}}{{\sqrt 7 }}) is correct because it represents a line that is tangent to both the hyperbola and the circle.
399 Medium
399 of 429
Analyzing +4 | -1
Two conics x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 and x2=−abyx^2=-\frac{a}{b} y intersect, if
View answer and explanation
Correct option: B
0<a<120 < a < \frac{1}{2}
Option A (0<a<120 < a < \frac{1}{2}) is correct because the condition for intersection requires the discriminant of the resulting quadratic equation to be positive, leading to the inequality 0<a<120 < a < \frac{1}{2}.
400 Medium
400 of 429
Analyzing +4 | -1
The circle drawn on the line segment joining the foci of the hyperbola x2a2−y2b2=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 as diameter cuts the asymptotes at
View answer and explanation
Correct option: D
(±a,±  b)\left( { \pm a, \pm \;b} \right)
Option C ((±  b,±  a)\left( { \pm \;b, \pm \;a} \right)) is correct because the circle intersects the asymptotes at these points.
401 Medium
401 of 429
Application +4 | -1
The locus of point of intersection of straight lines xa+yb=λ\frac{x}{a}+\frac{y}{b}=\lambda and xa−yb=1λ\frac{x}{a}-\frac{y}{b}=\frac{1}{\lambda} where λ\lambda is variable, is a
View answer and explanation
Correct option: C
Hyperbola
The correct answer is Hyperbola because multiplying the two given equations eliminates λ\lambda and results in the standard form of a hyperbola.
402 Medium
402 of 429
Application +4 | -1
Find the equation of the hyperbola whose foci are (6,4)(6,4) and (−4,4)(-4,4) and eccentricity is 2 .
View answer and explanation
Correct option: A
12x2−4y2−24x+32y−127=012 x^2-4 y^2-24 x+32 y-127=0
Option A (12x2−4y2−24x+32y−127=012 x^2-4 y^2-24 x+32 y-127=0) is correct because it represents the equation of the hyperbola derived from the given foci and eccentricity.
403 Easy
403 of 429
Understanding +4 | -1
If ee and e1e_1 are the eccentricities of a hyperbola and its conjugate, then 1e2+1e12\frac{1}{e^2}+\frac{1}{e_1^2} is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because 1e2+1e12=1\frac{1}{e^2} + \frac{1}{e_1^2} = 1 for a hyperbola and its conjugate.
404 Easy
404 of 429
Application +4 | -1
The tangents to the hyperbola x2−y2=3{x^2} - {y^2} = 3 are parallel to the straight line 2x+y+8=02x + y + 8 = 0 at the following points:
View answer and explanation
Correct option: B
(2,−1),(−2,1)\left( {2, - 1} \right),\left( { - 2,1} \right)
Option B ((2,−1),(−2,1)\left( {2, - 1} \right),\left( { - 2,1} \right)) is correct because these are the points where tangents parallel to the given line touch the hyperbola, calculated using the tangent's slope.
405 Easy
405 of 429
Application +4 | -1
If H(x,y)=0H(x, y)=0 represents the equation of a hyperbola and A(x,y)=0,C(x,y)=0A(x, y)=0, C(x, y)=0 the joint equation of its asymptotes and the conjugate hyperbola respectively, then for any point (α,β)(\alpha, \beta) in the plane, H(α,β),A(α,β)H(\alpha, \beta), A(\alpha, \beta), and C(α,β)C(\alpha, \beta) are in
View answer and explanation
Correct option: A
A.P.
Option A (A.P.) is correct because the relationship between the hyperbola, its asymptotes, and the conjugate hyperbola results in an arithmetic progression.

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406 Easy
406 of 429
Application +4 | -1
The line 42x−5y=404 \sqrt{2} x-5 y=40 touches the hyperbola x2100−y264=1\frac{x^2}{100}-\frac{y^2}{64}=1 at the point
View answer and explanation
Correct option: B
(102,8)(10 \sqrt{2}, 8)
Option B ((102,8)(10 \sqrt{2}, 8)) is correct because the point satisfies the equation of the tangent to the hyperbola at that point.
407 Easy
407 of 429
Application +4 | -1
The area of the triangle formed by the asymptotes and any tangent to the hyperbola x2−y2=a2x^2-y^2=a^2 is
View answer and explanation
Correct option: D
a2a^2
Option D is correct because the area of the triangle formed by the asymptotes and any tangent to the hyperbola is a2a^2.
408 Medium
408 of 429
Application +4 | -1
The centre of the hyperbola 9x2−16y2+18x+32y−151=09 x^2-16 y^2+18 x+32 y-151=0 is
View answer and explanation
Correct option: B
(−1,1)(-1,1)
Option B ((-1,1)) is correct because it is the center of the hyperbola found by completing the square and rewriting the equation in standard form.
409 Easy
409 of 429
Application +4 | -1
The line xcos⁡α+ysin⁡α=px \cos \alpha+y \sin \alpha=p touches the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, if
View answer and explanation
Correct option: A
a2cos⁡2α−b2sin⁡2α=p2a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2
Option A (a2cos⁡2α−b2sin⁡2α=p2a^2 \cos ^2 \alpha-b^2 \sin ^2 \alpha=p^2) is correct because it represents the condition for the line to be tangent to the hyperbola.
410 Medium
410 of 429
Application +4 | -1
The product of perpendiculars drawn from any point of a hyperbola to its asymptotes is
View answer and explanation
Correct option: A
a2b2a2+b2\frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}
Option A (a2b2a2+b2\frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}) is correct because the product of the perpendiculars from any point on a hyperbola to its asymptotes is a constant value, which is given by the formula a2b2a2+b2\frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}.
411 Easy
411 of 429
Application +4 | -1
If the chords of contact of the tangents from two points (x1,y1)\left(x_1, y_1\right) and (x2,y2)\left(x_2, y_2\right) to the hyperbola 4x2−5y2=a24 x^2-5 y^2=a^2 are at right angles. Then 16x1x2+25y1y216 x_1 x_2+25 y_1 y_2 is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the condition for perpendicularity of the chords of contact leads to the equation 16x1x2+25y1y2=016 x_1 x_2+25 y_1 y_2=0.
412 Medium
412 of 429
Application +4 | -1
Locus of the mid-point of the chords of the hyperbola x2−y2=a2x^2-y^2=a^2, that touch the parabola y2=4axy^2=4 a x, is
View answer and explanation
Correct option: B
y2(x−a)=x3y^2(x-a)=x^3
Option B (y2(x−a)=x3y^2(x-a)=x^3) is correct because it represents the locus of the mid-point of the chords of the hyperbola that touch the parabola.
413 Easy
413 of 429
Application +4 | -1
The equation of the hyperbola referred to its axes as axes of coordinates whose distance between the foci is 16 and eccentricity 2\sqrt{2}, is
View answer and explanation
Correct option: B
x2−y2=32x^2-y^2=32
Option B (x2−y2=32x^2-y^2=32) is correct because given the distance between foci and eccentricity, the equation of the hyperbola can be derived as x2−y2=32x^2-y^2=32.
414 Easy
414 of 429
Application +4 | -1
The equation to the hyperbola having its eccentricity 2 and the distance between its foci is 8, is
View answer and explanation
Correct option: B
x24−y212=1\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1
Option B (x24−y212=1\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1) is correct because it represents the equation of a hyperbola with the given eccentricity and distance between foci.
415 Easy
415 of 429
Application +4 | -1
The eccentricity of the hyperbola conjugate to x2−3y2=2x+8{x^2} - 3{y^2} = 2x + 8 is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the eccentricity of the conjugate hyperbola is calculated after finding the equation of the conjugate hyperbola.
416 Easy
416 of 429
Remembering +4 | -1
The curve xy=c2x y=c^2 is said to be
View answer and explanation
Correct option: B
A rectangular hyperbola
Option B (A rectangular hyperbola) is correct because the equation xy=c2xy = c^2 represents a rectangular hyperbola.
417 Easy
417 of 429
Application +4 | -1
The distance between the foci of the conic 7x2−9y2=637{x^2} - 9{y^2} = 63 is equal to
View answer and explanation
Correct option: A
8
Option A (8) is correct because it is the distance between the foci of the hyperbola, calculated using the formula 2ae.
418 Medium
418 of 429
Application +4 | -1
The equation of the tangents to the conic 3x2−y2=33 x^2-y^2=3 perpendicular to the line x+3y=2x+3 y=2 is
View answer and explanation
Correct option: A
y=3x±6y=3 x \pm \sqrt{6}
Option A (y=3x±6y=3 x \pm \sqrt{6}) is correct because it represents the tangent lines with a slope perpendicular to the given line, derived from the hyperbola equation.
419 Medium
419 of 429
Analyzing +4 | -1
The curve with parametric equation x=et+e−t  y=et−  e−tx = {e^t} + {e^{ - t}}\;y = {e^t} - \;{e^{ - t}} and is
View answer and explanation
Correct option: C
A hyperbola
Option C is correct because eliminating the parameter 't' results in the equation of a hyperbola.
420 Easy
420 of 429
Application +4 | -1
The diameter of 16x2−9y2=14416{x^2} - 9{y^2} = 144 which is conjugate to x=2yx = 2y is
View answer and explanation
Correct option: B
y=32x9y = \frac{{32x}}{9}
Option B (y=32x9y = \frac{{32x}}{9}) is correct because it represents the diameter conjugate to the given diameter of the hyperbola.
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