Question 1

If the eccentricity e of the hyperbola x^2 a^2 - y^2 b^2 = 1, passing through (6, 4 3 ), satisfies 15(e^2 + 1) = 34e, then the length of the latus rectum of the hyperbola x^2 b^2 - y^2 2(a^2+1) = 1 is

Accepted Answer

The given equation of the hyperbola is x^2 a^2 - y^2 b^2 = 1. Since it passes through (6, 4 3 ), we have: 36 a^2 - 48 b^2 = 1 The eccentricity e satisfies 15(e^2 + 1) = 34e. 15e^2 - 34e + 15 = 0 (3e - 5)(5e - 3) = 0 Since the eccentricity of a hyperbola is e > 1, we get e = 5 3 . We know that b^2 = a^2(e^2 - 1). b^2 = a^2 ( 25 9 - 1 ) = 16 9 a^2 Substituting b^2 into the first equation: 36 a^2 - 48 16 9 a^2 = 1 36 a^2 - 27 a^2 = 1 9 a^2 = 1 a^2 = 9 Then, b^2 = 16 9 9 = 16. The second hyperbola i

Question 2

Let the eccentricity e of a hyperbola satisfy the equation 6e^2 - 11e + 3 = 0. If the foci of the hyperbola are (3, 5) and (3, -4), then the length of its latus rectum is :

Accepted Answer

Given the equation for eccentricity: 6e^2 - 11e + 3 = 0 6e^2 - 9e - 2e + 3 = 0 (3e - 1)(2e - 3) = 0 e = 1 3 or e = 3 2 For a hyperbola, e > 1, so e = 3 2 . The distance between the foci (3, 5) and (3, -4) is 2ae. 2ae = (3-3)^2 + (5 - (-4))^2 = 9 2a ( 3 2 ) = 9 3a = 9 a = 3 Using the relation b^2 = a^2(e^2 - 1): b^2 = 9 ( 9 4 - 1 ) = 9 ( 5 4 ) = 45 4 The length of the latus rectum is 2b^2 a : 2 ( 45 4 ) 3 = 45 6 = 15 2 Answer: 15 2

Question 3

Let H: x^2 a^2 - y^2 b^2 =1 be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is 8 3 . If the line x= intersects the hyperbola H at the points A and

Accepted Answer

Given distance between foci is 2ae = 6 ae = 3 Distance between directrices is 2a e = 8 3 Multiplying the two equations, we get 4a^2 = 16 a^2 = 4 Dividing the two equations, we get e^2 = 9 4 Using b^2 = a^2(e^2 - 1), we get b^2 = 4 ( 9 4 - 1 ) = 5 The equation of the hyperbola is x^2 4 - y^2 5 = 1 The line x = intersects the hyperbola at A and B. Substituting x = , we get ^2 4 - y^2 5 = 1 y = 5( ^2 - 4) 2 The coordinates of A and B are ( , 5( ^2 - 4) 2 ) and ( , - 5( ^2 - 4) 2 ) The length of the

Question 4

Let O be the origin, and P and Q be two points on the rectangular hyperbola xy = 12 such that the mid point of the line segment PQ is ( 1 2 , - 1 2 ). Then the area of the triangle OPQ equals:

Accepted Answer

The equation of the chord of the hyperbola xy = 12 with midpoint (x_1, y_1) is given by T = S_1. x y_1 + y x_1 2 - 12 = x_1 y_1 - 12 x y_1 + y x_1 = 2 x_1 y_1 Substituting the midpoint ( 1 2 , - 1 2 ): x (- 1 2 ) + y ( 1 2 ) = 2 ( 1 2 ) (- 1 2 ) -x + y = -1 y = x - 1 To find the coordinates of P and Q, substitute y = x - 1 into the equation of the hyperbola xy = 12: x(x - 1) = 12 x^2 - x - 12 = 0 (x - 4)(x + 3) = 0 x = 4, -3 For x = 4, y = 3. For x = -3, y = -4. Thus, the coordinates of P and Q

Question 5

For some (0, 2 ), let the eccentricity and the length of the latus rectum of the hyperbola x^ 2 -y^ 2 ^ 2 =8 be e_ 1 and l_ 1 , respectively, and let the eccentricity and the length of the latus rectu

Accepted Answer

x^2 8 - y^2 8 ^2 = 1, e_1 = 1 + 8 ^2 8 _1 = 2b^2 a = 2 (8 ^2 ) 2 2 x^2 6 + y^2 6 ^2 = 1; e_2 = 1 - 6 ^2 6 = _2 = 2b^2 a = 2 6 ^2 6 e_1^2 = e_2^2(1 + ^2 ) 1 + ^2 = ^2 (1 + 1 ^2 ) 1 + ^2 = ^2 + ^2 Solving we get = 4 _1 = 2 2 e_1 = 3 2 _2 = 6 e_2 = 1 2 ( _1 _2 e_1 e_2 ) ^2 = 8 (By putting values)

Question 6

Let PQ be a chord of the hyperbola x^ 2 4 - y^ 2 b^ 2 =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 hyperbo

Accepted Answer

Hyperbola x^2 4 - y^2 b^2 = 1 with e = 3 . e^2 = 1 + b^2 4 = 3 b^2 = 8. Let P = (x_0, y_0), Q = (x_0, -y_0). For equilateral OPQ: OP = PQ x_0^2 + y_0^2 = 4y_0^2 x_0^2 = 3y_0^2. From the hyperbola: 3y_0^2 4 - y_0^2 8 = 1 5y_0^2 8 = 1 y_0^2 = 8 5 . Area = 3 4 (2y_0)^2 = 3 y_0^2 = 8 3 5 .

Question 7

Let the domain of the function f(x)= _ 3 _ 5 _ 7 (9 x-x^ 2 -13 ) be the interval ( m , n ). Let the hyperbola x^ 2 a ^ 2 - y^ 2 ~b ^ 2 =1 have eccentricity n 3 and the length of the latus rectum 8 ~m

Accepted Answer

For f(x) = _3 _5 _7(9x-x^2-13): _7( ) > 0 9x-x^2-13 > 1 (x-2)(x-7) _5( ) > 0 _7(9x-x^2-13) > 1 9x-x^2-13 > 7 (x-4)(x-5) Domain = (4, 5), so m = 4, n = 5. Eccentricity e = n/3 = 5/3, latus rectum = 8m/3 = 32/3. e^2 = 1 + b^2/a^2 b^2/a^2 = 16/9. 2b^2/a = 32/3 32a/9 = 32/3 a = 3, so a^2 = 9, b^2 = 16. b^2 - a^2 = 16 - 9 = 7.

Question 8

Let P (10,2 15 ) be a point on the hyperbola x^ 2 a ^ 2 - y^ 2 ~b ^ 2 =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 :

Accepted Answer

For the hyperbola x^2 a^2 - y^2 b^2 = 1, point P(10, 2 15 ) gives 100 a^2 - 60 b^2 = 1. The latus rectum length 2b^2 a = 8 gives b^2 = 4a. Substituting: 100 a^2 - 15 a = 1, which yields a^2 + 15a - 100 = 0, so a = 5 and b^2 = 20. Thus c^2 = 45 and c = 3 5 . The foci are S( 3 5 , 0). Triangle PSS' has base SS' = 6 5 and height 2 15 . Area = 1 2 6 5 2 15 = 30 3 . Therefore (Area)^2 = 2700.

Question 9

If the line x+2 y=1, where R , does not meet the hyperbola x^ 2 -9 y^ 2 =9, then a possible value of is:

Accepted Answer

Hyperbola: x^2 9 - y^2 = 1. Line: y = 1 - x 2 . Substituting: x^2 - 9 (1- x)^2 4 = 9 (4 - 9 ^2)x^2 + 18 x - 45 = 0. For no intersection (when 4 - 9 ^2 0): discriminant D = 324 ^2 + 180(4 - 9 ^2) -1296 ^2 + 720 5 9 . | | > 5 3 0.745. Among the options, only = 0.8 satisfies this condition.

Question 10

Let the foci of a hyperbola coincide with the foci of the ellipse x^ 2 36 + y^ 2 16 =1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

Accepted Answer

For ellipse x^2 36 + y^2 16 = 1: a = 6, b = 4, c = 36-16 = 2 5 . Foci of ellipse: ( 2 5 , 0). For hyperbola with eccentricity e = 5 and foci at ( 2 5 , 0): e = C A A = 2 5 5 . C^2 = A^2 + B^2 20 = 4 5 + B^2 B^2 = 96 5 . Latus rectum = 2B^2 A = 2 96 5 2 5 5 = 96 5 .

Hyperbola — Previous Year Questions & Solutions