Question

Find the equation of the tangent to the hyperbola $4x^2-9y^2=1$, which is parallel to the line $4y=5x+7$

• A. $48y=60x\pm \sqrt{161}$
• B. $24y=30x\pm \sqrt{147}$
• C. $48y=60x\pm \sqrt{147}$
• D. $24y=30x\pm \sqrt{161}$

Answer 1

The correct option is D $24y=30x\pm \sqrt{161}$

For the hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Equation of tangent of slope m is given by

$y=mx\pm \sqrt{m^2{a}^2-b^2}$

For the given hyperbola

$a=\frac{1}{2},b=\frac{1}{3}$

Equation of tangent is

$y=mx\pm \sqrt{\frac{m^2}{4}-\frac{1}{9}}$

As, tangent is parallel to the line $4y=5x+7$

So, $m=\frac{5}{4}$

Hence the equation of tangent becomes

$y=\frac{5}{4}x\pm \sqrt{\frac{25}{4\times 16}-\frac{1}{9}}$

$y=\frac{5}{4}x\pm \sqrt{\frac{161}{9\times 64}}$

$24y-30x=\pm \sqrt{161}$