MereTeacher

JEE • NEET • FOUNDATION

JEE Previous Year Question

$The equation of the hyperbola whose foci are (-2, 0) and (2, 0) and eccentricity is 2 is given by$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $x2 - 3y2 = 3$
B $3x2 - y2 = 3$
C $- x2 + 3y2 = 3$
D $- 3x2 + y2 = 3$

Correct answer: B

Solution

(b) ae = 2 e = 2 ∴ a = 1 b2 = a2 (e2 - 1) b2 = 1 (4 - 1) b2 = 3 \frac{x ^{2}}{1} - \frac{y ^{2}}{3} = 1 3x2 - y2 = 3

More Hyperbola PYQs

$The length of the chord of the parabola x^{2} = 4 y havingequation x - 2 y + 42 = 0 is:$
$The equation of the pair of asymptotes of the hyperbola x^{2} - 2 y^{2} - 8 x + 8 y + 4 = 0 is$
$Suppose that the circle x^{2} + y^{2} + 2 gx + 2 fy + c = 0 has its centre on 2 x + 3 y - 7 = 0 and cuts the circles\dots$
$The product of lengths of perpendicular from any point on the hyperbola x^{2} - y^{2} = 16 to its asymptotes, is$
$If x^{2} = 8 ay is the transformed equation of x^{2} - 4 y + 6 x + 15 = 0 when the origin is shifted to the point\dots$