Question

x + 2y -1=0 and 3x+2y+5=0 are asymptote of a hyperbola which is passing through (2,3) . Find equation of hyperbola amd its conjugate.​

Answer 1

Answer:

$\bold\green{Hyperbola:}$

$\bold\red{3{x}^{2}+4{y}^{2}+8xy+2x+8y-124=0}$

$\bold\green{Conjugate\:Hyperbola:}$

$\bold\red{3{x}^{2}+4{y}^{2}+8xy+2x+8y+114=0}$

Step-by-step explanation:

Given,

x + 2y -1=0 and 3x+2y+5=0 are asymptotes of a hyperbola.

Therefore,

pair of asymptotes can be written as,

(x+2y-1)(3x+2y+5)=0

Now,

Let the equation of Hyperbola is ,

(x+2y-1)(3x+2y+5)+ μ =0

where,

μ is a parameter.

Now, we have to find the value of μ.

It is given that,

Hyperbola is passing through the pt. (2,3)

Therefore,

it will satisfy the assumed equation of Hyperbola.

Therefore, putting the values,

we get,

=> (2+6-1)(6+6+5) + μ = 0

=> 7 × 17 +μ = 0

=> μ + 119 = 0

=> μ = -119

Thus,

the equation of Hyperbola is

(x+2y-1)(3x+2y+5) - 119 =0

Therefore,

the equation of it's conjugate is

(x+2y-1)(3x+2y+5) + 119 =0

Hence,

Hyperbola:

$\bold{3{x}^{2}+4{y}^{2}+8xy+2x+8y-124=0}$

Conjugate Hyperbola:

$\bold{3{x}^{2}+4{y}^{2}+8xy+2x+8y+114=0}$