Mathematics Objective Questions and Answers: Hyperbola
Review structured MCQ sets for JEE Mathematics Hyperbola. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Download Hyperbola MCQs with Answers
Access the complete set of multiple-choice questions for Hyperbola below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.
Question: The eccentricity of the conic x2 – y2 – 4x + 4y + 16 = 0 is
- a) √2
- b) 1
- c) 2
- d) 1/2
Answer: √2
Question: The equation 9x2 – 16y2 – 18x + 32y – 151 = 0 represent a hyperbola -
- a) Equation of whose directrix is x = 21/5 and x = -11/5
- b) None of these
- c) Length of whose transverse axes is 4
- d) Length of whose latusrectum is 9
Answer: Equation of whose directrix is x = 21/5 and x = -11/5
Question: For what value of λ does the line y = 2x + λ touches the hyperbola 16x2 – 9y2 = 144 ?
- a) ± 2 √5
- b) ± 3 √5
- c) ± 4 √5
- d) None of these
Answer: ± 2 √5
Question: Find the equation of the tangent to the hyperbola x2 – 4y2 = 36 which is perpendicular to the line x – y + 4 = 0.
- a) x + y ± 3 √3 = 0
- b) x + y ± 2 √3 = 0
- c) x + y ± 5 √3 = 0
- d) x – y ± 3 √3 = 0
Answer: x + y ± 3 √3 = 0
Question: If the normals at (xi , yi ) i = 1, 2, 3, 4 to the rectangular hyperbola xy = 2 meet at the point (3, 4), then
(1) x1 + x2 + x3 + x4 = 3 (2) y1 + y2 + y3 + y4 = 4
(3) x1 x2 x3 x4 = –4 (4) y1 y2 y3 y4 = 4
- a) 1, 2 and 3 are correct
- b) 2 and 4 are correct
- c) 1 and 2 are correct
- d) 1 and 3 are correct
Answer: 1, 2 and 3 are correct
Question: If the circle x2 + y2 = 1 cuts the rectangular hyperbola xy = 1 in four points ( xi , yi ); i = 1, 2, 3, 4 then.
(1) x1x2 x3x4 = – 1 (2) y1y2 y3 y4 = 1
(3) x1 + x2 + x3 + x4 = 1 (4) y1 + y2 + y3 + y4 = 0
- a) 2 and 4 are correct
- b) 1 and 2 are correct
- c) 1, 2 and 3 are correct
- d) 1 and 3 are correct
Answer: 2 and 4 are correct
Question:
Statement-1 : There can be infinite points from where we can draw two mutually perpendicular tangents to the hyperbola x2/9 - y2/16 =1
atement-2 : The director circle in case of hyperbola x2/9 - y2/16 =1will not exist because a2 < b2 and director circle is x2 + y2 = a2 – b2.
- a) Statement-1 is False, Statement-2 is True.
- b) Statement-1 is True, Statement-2 is False.
- c) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
- d) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
Answer: Statement-1 is False, Statement-2 is True.
Question:
Statement-1 : With respect to a hyperbola x2/9 - y2/16 =1 perpendicular are drawn from a point (5, 0) on the lines 3y ± 4x = 0, then their feet lie on circle x2 + y2 = 16.
Statement-2 : If from any foci of a hyperbola perpendicular are drawn on the asymptotes of the hyperbola then their feet lie on auxiliary circle.
- a) Statement-1 is False, Statement-2 is True.
- b) Statement-1 is True, Statement-2 is False.
- c) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
- d) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
Answer: Statement-1 is False, Statement-2 is True.
More Questions....................................
Question: Two straight lines pass through the fixed points (± a,0) and have gradients whose product is k, then the locus of the points of inter-section of the lines is
- a) hyperbola
- b) parabola
- c) circle
- d) None of these
Answer: hyperbola
Question: The eccentricity of the conic x2 – y2 – 4x + 4y + 16 = 0 is
- a) √2
- b) 1/2
- c) 2
Answer: √2
Question: The equation 16x2 = 3y2 – 32x + 12y – 44= 0 represents a hyperbola
- a) whose eccentricity is √19/3
- b) whose centre is (– 1, 2)
- c) the length of whose conjugate axis is 4
- d) the length of whose transverse axis is 4√3
Answer: whose eccentricity is √19/3
Question: The line 5x + 12y = 9 touches the hyperbola x2 – 9y2 = 9 at the point
- a) (5, – 4/3)
- b) None of these
- c) (– 5,4/3)
- d) (3, – 1/2)
Answer: (5, – 4/3)
Question: Locus of the mid points of the chords of the circle x2 + y2 = 16, which are tangent to the hyperbola 9x2 – 16y2 = 144 is
- a) (x2 + y2)2 = 16x2 – 9y2
- b) (x2 + y2)2 = 5x2 – 4y2
- c) (x2 + y2)2 = 19x2 – 8y2
- d) (x2 + y2)2 = 18x2 – 8y2
Answer: (x2 + y2)2 = 16x2 – 9y2
Question: If e and e' be the eccentricities of a hyperbola and its conjugate then the value of 1/e2 + 1/e' 2 =
- a) 1
- b) 4
- c) 0
- d) 2
Answer: 1
Question: The equation of the common tangents to the parabola y2 = 8x and the hyperbola 3x2 – y2 = 3 is-
- a) 2x ± y + 1 = 0
- b) x ± 2y + 1 = 0
- c) x ± y + 1 = 0
- d) x ± y + 2 = 0
Answer: 2x ± y + 1 = 0
Question: The locus of the point of intersection of the lines √3 x –y– 4 √3 k =0 and √3 kx + ky – 4 √3 = 0 for different values of k is-
- a) Hyperbola
- b) Parabola
- c) Ellipse
- d) Circle
Answer: Hyperbola
Question: The area of a triangle formed by the lines x – y = 0, x + y = 0 and any tangent to the hyperbola x2 – y2 = a2 is-
- a) a2
- b) 3a2
- c) 2a2
- d) 4a2
Answer: a2
Question: Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1,2) and eccentricity √3 .
- a) 7x2 – 2y2 + 12xy – 2x + 14y – 22 = 0
- b) 7x2 – 2y2 – 12xy – 2x + 14y – 18 = 0
- c) 7x2 + 2y2 – 12xy – 2x + 14y – 22 = 0
- d) 7x2 + 2y2 + 12xy – 2x + 14y + 11 = 0
Answer: 7x2 – 2y2 + 12xy – 2x + 14y – 22 = 0
Question: From the points on the circle x2 + y2 = a2, tangents are drawn to the hyperbola x2 – y2 = a2; then the locus of the middle points of the chords of contact is the curve
- a) (x2 – y2)2 = a2(x2 + y2)
- b) (x2 – y2)2 = a2(x2 – y2)
- c) (x2 + y2)2 = a2(x2 + y2)
- d) (x2 – y2)2 = 2a2(x2 + y2)
Answer: (x2 – y2)2 = a2(x2 + y2)
Question: Coordinates of foci, eccentricity of the hyperbola 9x2 – 16y2 – 72x + 96y– 144 = 0 are
- a) (9,3), (– 1,3), 5/4
- b) (3,3), (– 1,3), 1/4
- c) (1,3), (– 2,3), 5/4
- d) (2,3), (–2, 3), 3/4
Answer: (9,3), (– 1,3), 5/4
Question: The locus of the mid point of the chords of the circle x2 + y2 = 16, which are tangent to the hyperbola 9x2 – 16y2 = 144 is-
- a) (x2 + y2)2 = a2x2 – b2y2
- b) x2 + y2 = a2 – b2
- c) (x2 + y2)2 = a2 – b2
- d) (x2 + y2)2 = a2 + b2
Answer: (x2 + y2)2 = a2x2 – b2y2
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JEE Mathematics Hyperbola Objective Test Questions
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FAQs
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