If the eccentricity e of the hyperbola x2a2- y2b2=1, passing through (6, 4 3 ), satisfies 15 (e2+1 )=34e, then the length of the latus rectum of the hyperbola x2b2- y22 (a2+1 )=1 is :
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Previous year questions with solutions · 2025 – 2026
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If the eccentricity e of the hyperbola x2a2- y2b2=1, passing through (6, 4 3 ), satisfies 15 (e2+1 )=34e, then the length of the latus rectum of the hyperbola x2b2- y22 (a2+1 )=1 is :
The eccentricity of an ellipse E with centre at the origin O is 32 and its directrices are x= 4 63. Let H: x2a2- y2b2=1 be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose length of latus rectum is equal to the length of minor axis of E. Then the distance between the foci of H is :
Let the eccentricity e of a hyperbola satisfy the equation 6e2-11e+3=0. If the foci of the hyperbola are (3, 5) and (3, -4), then the length of its latus rectum is :
Let H : x2a2- y2b2=1 be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is 83. If the line x= intersects the hyperbola H at the points A and B such that the area of the triangle AOB is 4 15, where O is the origin, then 2 equals
Let O be the origin, and P and Q be two points on the rectangular hyperbola xy=12 such that the midpoint of the line segment PQ is ( 12, - 12 ). Then the area of the triangle OPQ equals :
For some (0, 2 ), let the eccentricity and the length of the latus rectum of the hyperbola x2 - y2 2 = 8 be e1 and l1, respectively, and let the eccentricity and the length of the latus rectum of the ellipse x2 2 + y2 = 6 be e2 and l2, respectively. If e12 = e22 ( 2 + 1 ), then ( l1 l2e1 e2 ) 2 is equal to .
Let the ellipse E : x2144 + y2169 = 1 and the hyperbola H : x216 - y2 2 = -1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, then the value of 24(e + L) is :
Let the domain of the function f(x) = 3 5 7 (9x - x2 - 13 ) be the interval (m, n). Let the hyperbola x2a2 - y2b2 = 1 have eccentricity n3 and the length of the latus rectum 8m3. Then b2 - a2 is equal to :
Let PQ be a chord of the hyperbola x24 - y2b2 = 1, perpendicular to the x-axis such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is 3, then the area of the triangle OPQ is
Let P(10, 2 15) be a point on the hyperbola x2a2 - y2b2 = 1, whose foci are S and S'. If the length of its latus rectum is 8, then the square of the area of PSS' is equal to :
Let the foci of a hyperbola coincide with the foci of the ellipse x236 + y216 = 1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :
Consider the hyperbola x2a2 - y2b2 = 1 having one of its foci at P(-3, 0). If the latus rectum through its other focus subtends a right angle at P and a2 b2 = 2 - , , N, then + is .
Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be (-5, 0) and 5x + 9 = 0, respectively. If the product of the focal distances of a point ( , 2 5) on the hyperbola is p, then 4p is equal to .
Let the sum of the focal distances of the point P(4, 3) on the hyperbola H: x2a2 - y2b2 = 1 be 8 53. If for H, the length of the latus rectum is l and the product of the focal distances of the point P is m, then 9l2 + 6m is equal to :
Let the product of the focal distances of the point P(4, 2 3) on the hyperbola H: x2a2 - y2b2 = 1 be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then p2 + q2 is equal to
If the equation of the hyperbola with foci (4, 2) and (8, 2) is 3x2 - y2 - x + y + = 0, then + + is equal to .
Let one focus of the hyperbola H : x2a2 - y2b2 = 1 be at ( 10, 0) and the corresponding directrix be x = 9 10. If e and l respectively are the eccentricity and the length of the latus rectum of H, then 9(e2 + l) is equal to :