Find the equation of hyperbola whose vertices at ( ± 6 , 0 ) and foci at ( ± 8 , 0 ) .

Given that, we have vertices ( ± 6 , 0 ) and foci ( ± 8 , 0 ) . Since the vertices and foci are of the form ± a , 0 and ± c , 0 Hence, it is a horizontal hyperbola. So, the equation of hyperbola is of the form x 2 a 2 - y 2 b 2 = 1 Given vertices of hyperbola ( ± 6 , 0 ) ∴ a = 6 ∴ a 2 = 36 → 1 And Foci is of the form ( ± c , 0 ) ∴ c = 8 Now, c 2 = a 2 + b 2 ⇒ 64 = 36 + b 2 ⇒ b 2 = 64 - 36 ∴ b 2 = 28 → 2 Put 1 and 2 in standard hyperbola, so we get: x 2 36 - y 2 28 = 1 Hence, the required equation of the hyperbola is x 2 36 - y 2 28 = 1 .

Find the equation of the hyperbola with vertices at ( 0 , ± 5 ) and foci at ( 0 , ± 8 ) .

Given that, the vertices = ( 0 , ± 5 ) and foci = ( 0 , ± 8 ) The vertices are of the form ( 0 , ± a ) Therefore, a = 5 ⇒ a 2 = 25 Thus, the required equation of the hyperbola is y 2 a 2 - x 2 b 2 = 1 And foci are of the form 0 , ± c ∴ c = 8 Now, c 2 = a 2 + b 2 ⇒ b 2 = c 2 - a 2 ⇒ b 2 = 8 2 - 5 2 ⇒ b 2 = 64 - 25 ⇒ b 2 = 39 Thus, b 2 = 39 , a 2 = 25 Hence, the required equation of the hyperbola is y 2 25 - x 2 39 = 1 .

Find the equation of hyperbola whose foci are ( ± 29 , 0 ) and the transverse axis is of the length 10 units.

Given that the foci are ± 29 , 0 The foci are of the form ± c , 0 ∴ c = 29 Thus, the required equation of the hyperbola is x 2 a 2 - y 2 b 2 = 1 And length of transverse axis is 2 a = 10 units ∴ a = 5 Now, c 2 = a 2 + b 2 ⇒ b 2 = c 2 - a 2 ⇒ b 2 = 29 - 25 ⇒ b 2 = 4 Thus, b 2 = 4 , a 2 = 25 Hence, the required equation of the hyperbola is x 2 25 - y 2 4 = 1 .

Find the equation of hyperbola whose foci are ( ± 5 , 0 ) and the conjugate axis is of the length 8 units. Also, find its eccentricity.

Given that, foci = ( ± 5 , 0 ) And it is of the form ± c , 0 ∴ c = 5 Therefore, the required equation of the hyperbola is x 2 a 2 - y 2 b 2 = 1 And length of conjugate axis is 8 units ∴ 2 b = 8 ⇒ b = 4 Now, c 2 = a 2 + b 2 ⇒ a 2 = c 2 - b 2 ⇒ a 2 = 25 - 16 ⇒ a 2 = 9 Thus, b 2 = 16 , a 2 = 9 Hence, the required equation of the hyperbola is x 2 9 - y 2 16 = 1 So the eccentricity of the hyperbola is e = c a Thus, e = 5 3 .

Find the equation of hyperbola whose foci are ( ± 3 5 , 0 ) and the the length of latus rectum is 8 units.

Given that, foci = ( ± 3 5 , 0 ) And it is of the form ± c , 0 . ∴ c = 3 5 Thus, the required equation of hyperbola is x 2 a 2 - y 2 b 2 = 1 And the length of latus-rectum is 8 units ∴ 2 b 2 a = 8 ⇒ b 2 = 4 a Now, c 2 = a 2 + b 2 ⇒ a 2 = 3 5 2 - 4 a ⇒ a 2 = 45 - 4 a ⇒ a 2 - 45 + 4 a = 0 ⇒ a 2 + 9 a - 5 a - 45 = 0 ⇒ a a + 9 - 5 a + 9 = 0 ⇒ a - 5 a + 9 = 0 ∴ a = 5 or a = - 9 ∴ a = 5 ( ' a ' cannot be negative) Since we know, b 2 = 4 a ∴ b 2 = 4 × 5 = 20 Thus, b 2 = 20 , a 2 = 25 Hence, the required equation of hyperbola is x 2 25 - y 2 20 = 1 .

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 .

E M B I B E

Senior Secondary School Mathematics FOR CLASS 11>Chapter 24 - Hyperbola>EXERCISE>Q 8

EASY

11th CBSE

IMPORTANT

Earn 100

Find the lengths of axes, coordinates of the vertices, coordinates of the foci, eccentricity and length of the latus-rectum of the following hyperbola:

$\frac{y^{2}}{9} - \frac{x^{2}}{27} = 1$ .

Important Questions on Hyperbola

EASY

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 24 - Hyperbola>EXERCISE>Q 9

Find the lengths of axes, coordinates of the vertices, coordinates of the foci, eccentricity and length of the latus-rectum of the following hyperbola:

$3 y^{2} - x^{2} = 108$ .

EASY

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 24 - Hyperbola>EXERCISE>Q 10

Find the lengths of axes, coordinates of the vertices, coordinates of the foci, eccentricity and length of the latus-rectum of the following hyperbola:

$5 y^{2} - 9 x^{2} = 36$ .

EASY

11th CBSE

IMPORTANT

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

Find the equation of hyperbola whose vertices at