Find the equation of hyperbola whose vertices are ( ± 2 , 0 ) and the eccentricity is 2 .

Since vertices are of the form ( ± a , 0 ) , so the required equation of hyperbola is x 2 a 2 - y 2 b 2 = 1 But, it is given that the vertices are ( ± 2 , 0 ) ∴ a = 2 And also, given that, the eccentricity is 2 ∴ c a = 2 ⇒ c = 2 a ⇒ c 2 = 4 a 2 Now, c 2 = a 2 + b 2 ⇒ 4 a 2 = 4 + b 2 ⇒ 4 2 2 = 4 + b 2 ∵ a = 2 ⇒ 16 = 4 + b 2 ⇒ b 2 = 12 Thus, b 2 = 12 , a 2 = 4 Hence, the required equation of the hyperbola is x 2 4 - y 2 12 = 1 .

Find the equation of hyperbola whose foci are ( ± 5 , 0 ) and the eccentricity is 5 3 .

Since the y - coordinate of the foci is 0 , so the required equation of hyperbola is x 2 a 2 - y 2 b 2 = 1 And its foci is ± c , 0 . But, it is given that the foci are ( ± 5 , 0 ) ∴ c = 5 And also, given that, the eccentricity is 5 3 ∴ c a = 5 3 ⇒ c = 5 3 a ⇒ c 2 = 5 3 a 2 ⇒ 5 2 = 5 3 a 2 ∴ a 2 = 3 Now, c 2 = a 2 + b 2 ⇒ 5 = 3 + b 2 ∵ a 2 = 3 ⇒ 2 = b 2 Thus, b 2 = 2 , a 2 = 3 Hence, the required equation of hyperbola is x 2 3 - y 2 2 = 1 .

Find the equation of the hyperbola with eccentricity 2 and the distance between whose foci is 16 .

Given the eccentricity of hyperbola is e = 2 . And also the distance between foci is 16 ∴ 2 a e = 2 c = 16 ⇒ 2 × a × 2 = 16 ⇒ a 2 = 8 ⇒ a 2 = 32 Now, c 2 = a 2 + b 2 ⇒ 8 2 = 32 + b 2 ∵ c = 8 , a 2 = 32 ⇒ 64 = 32 + b 2 ⇒ b 2 = 32 Thus, a 2 = 32 , b 2 = 32 . Hence, the equation of hyperbola is x 2 32 - y 2 32 = 1 .

Find the equation of the hyperbola whose vertices are ( 0 , ± 3 ) and the eccentricity is 4 3 . Also find the coordinates of its foci.

Since the vertices of the given hyperbola are of the form ( 0 , ± a ) , it is a case of vertical hyperbola. Let its equation be y 2 a 2 - x 2 b 2 = 1 Given vertices are ( 0 , ± 3 ) ∴ a = 3 And also the eccentricity is 4 3 ∴ e = c a = 4 3 ⇒ c = 4 × a 3 ⇒ c = 4 × 3 3 ⇒ c = 4 Now, c 2 = a 2 + b 2 ⇒ 4 2 = 3 2 + b 2 ⇒ 4 2 - 3 2 = b 2 ⇒ 7 = b 2 Thus, a 2 = 9 , b 2 = 7 Hence, the equation of hyperbola is y 2 9 - x 2 7 = 1 So, the foci of hyperbola is = 0 , ± c = 0 , ± 4 .

Find the equation of the hyperbola whose foci are ( 0 , ± 13 ) and the length of whose conjugate axis is 24 units.

Since the foci of the given hyperbola are of the form ( 0 , ± c ) , it is a case of vertical hyperbola. Let its equation be y 2 a 2 - x 2 b 2 = 1 Given foci are ( 0 , ± 13 ) ∴ c = 13 Given that, the length of the conjugate axis is 24 units. ∴ 2 b = 24 ⇒ b = 12 We know that, c 2 = a 2 + b 2 ⇒ 13 2 = a 2 + 12 2 ∵ c = 13 , b = 12 ⇒ 169 - 144 = a 2 ⇒ 25 = a 2 Thus, a 2 = 25 , b 2 = 144 Hence, the equation of hyperbola is y 2 25 - x 2 144 = 1 .

Find the equation of the hyperbola whose foci are ( 0 , ± 10 ) and the length of whose latus rectum is 9 units.

Since the foci of the given hyperbola are of the form ( 0 , ± c ) , it is a case of vertical hyperbola. Let its equation be y 2 a 2 - x 2 b 2 = 1 Given foci are ( 0 , ± 10 ) ∴ c = 10 Given that, the length of latus-rectum is 9 units. ∴ 2 b 2 a = 9 ⇒ 2 b 2 = 9 a ⇒ b 2 = 9 2 a We know that, c 2 = a 2 + b 2 ⇒ 10 2 = a 2 + 9 2 a ∵ c = 10 , b 2 = 9 2 a ⇒ 100 = a 2 + 9 2 a ⇒ 200 = 2 a 2 + 9 a ⇒ 2 a 2 + 9 a - 200 = 0 ⇒ a = - 9 ± 9 2 - 4 × 2 × - 200 2 × 2 [Using x = - b ± b 2 - 4 a c 2 a ] ⇒ a = - 9 ± 81 + 1600 4 ⇒ a = - 9 ± 1681 4 ⇒ a = - 9 ± 41 4 ⇒ a = - 9 + 41 4 or a = - 9 - 41 4 ⇒ a = 32 4 [Since a cannot be negative] ∴ a = 8 and we know, b 2 = 9 2 a ⇒ b 2 = 9 2 × 8 = 36 Thus, a 2 = 64 , b 2 = 36 Hence, the equation of hyperbola is y 2 64 - x 2 36 = 1 .

Find the equation of the hyperbola having its foci at ( 0 , ± 14 ) and passing through the point P ( 3 , 4 ) .

Since the foci of the given hyperbola are of the form ( 0 , ± c ) , so it is a case of vertical hyperbola. Let its equation be y 2 a 2 - x 2 b 2 = 1 → 1 Given foci are ( 0 , ± 14 ) ∴ c = 14 We know that, c 2 = a 2 + b 2 ⇒ 14 2 = a 2 + b 2 ⇒ 14 = a 2 + b 2 ⇒ b 2 = 14 - a 2 → 2 Since equation 1 passing through P ( 3 , 4 ) , then 4 2 a 2 - 3 2 b 2 = 1 ⇒ 16 a 2 - 9 b 2 = 1 ⇒ 16 a 2 - 9 14 - a 2 = 1 (using 2 ) ⇒ 16 14 - a 2 - 9 a 2 = a 2 14 - a 2 ⇒ 224 - 16 a 2 - 9 a 2 = 14 a 2 - a 4 ⇒ 224 - 25 a 2 = 14 a 2 - a 4 ⇒ 224 = 39 a 2 - a 4 ⇒ a 4 - 39 a 2 + 224 = 0 ⇒ a 4 - 32 a 2 - 7 a 2 + 224 = 0 ⇒ a 2 a 2 - 32 - 7 a 2 - 32 = 0 ⇒ a 2 - 32 = 0 or a 2 - 7 = 0 ∴ a 2 = 32 or a 2 = 7 Since we know, 14 = a 2 + b 2 If a 2 = 32 , then b 2 = - 18 (Neglected, because b 2 cannot be negative) If a 2 = 7 , then b 2 = 7 Thus, a 2 = 7 , b 2 = 7 Hence, the equation of hyperbola is y 2 7 - x 2 7 = 1 .

E M B I B E

Senior Secondary School Mathematics FOR CLASS 11>Coordinate Geometry>Chapter 24 - Hyperbola>EXERCISE

R S Aggarwal Solutions for Exercise 1: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 1: EXERCISE

Attempt the free practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. Senior Secondary School Mathematics FOR CLASS 11 solutions are prepared by Experienced Embibe Experts.