Find the equation of conjugate hyperbola whose equations of asymptotes are x + 2y + 3 = 0 and 3x + 4y + 5 = 0 and hyperbola passes through (1, -1).

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Solution

The correct option is B

Combined equation of asymptotes (x+2y+5)(3x+4y+5)=0

$3 x^{2} + 4 x y + 5 x + 6 x y + 8 y^{2} + 10 y + 9 x + 12 y + 15 = 0$

$3 x^{2} + 8 y^{2} + 10 x y + 14 x + 22 y + 15 = 0$

only difference between equation of hyperbola and the equation of asymptotes is constant part.

equation of hyperbola is

$3 x^{2} + 8 y^{2} + 10 x y + 14 x + 22 y + λ = 0$ (1)

Since it passes through the point(1,-1)

substituting the x=1 & y=-1 in equation (1)

$3 \times 1 + 8 \times 1 + 10 \times (1) (- 1) + 14 \times 1 + 22 \times (- 1) + λ = 0$

$3 + 8 - 10 + 14 - 22 y + λ = 0$

$λ = 7$

equation of hyperpola is

$3 x^{2} + 8 y^{2} + 10 x y + 14 x + 22 y + 7 = 0$

Constant part of equation of hyperbola,asymptotes and conjugate hyperbola are in AP

we observe that constant part of hyperbola is 7,asymptotes is 15.

then constant part of conjugate hyperpola is 23

Equation of conjugate hyperbola is

$3 x^{2} + 8 y^{2} + 10 x y + 14 x + 22 y + 23 = 0$

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