Question

A hyperbola passes through (1,2) and has asymptotes as x+ y-2=0 and 7x+y-8 0. Eccentricity of the hyperbola is equal to

Answer 1

Answer:

e = √10/3

Step-by-step explanation:

Hi,

We now that if 'a' length of transverse axis and 'b' the length of conjugate axis,

Acute angle between asymptotes is given by 2tan⁻¹(b/a).

Let ∅ be the angle between asymptotes,

Given asymptotes are x + y = 2 and 7x + y = 8

So, tan∅ = |-1+7|/1+7 = 6/8 = 3/4.

But ∅ = 2tan⁻¹(b/a)

=>tan∅/2 = b/a

=let tan∅/2 = x

=> tan∅ = 2tan(∅/2)/1-tan²(∅/2)

=> 2x/1-x² = 3/4

=> 8x = 3 - 3x²

=>3x² + 8x - 3 = 0

=>(3x-1)(x+3)=0

=> x = 1/3 or -3

But x = tan∅/2 = b/a which is always positive

Hence,

b/a = 1/3

Eccentricity of hyperbola , e = √1 +(b/a)²

=> e = √1 + 1/9

=> e = √10/9

=> e = √10/3

Hope, it helped !