Question

The equation of tangents to the curve $9y^2-4x^2=1$ which is parallel to the line $3y=x+7$ is/are

• A. $2x-6y=\sqrt{3}$
• B. $2x-6y=-\sqrt{3}$
• C. $16x-20y=25$
• D. $16x-20y=-25$

Answer 1

Answer: (A), (B)

$2x-6y=-\sqrt{3}$

Equation of tangent to a conjugate hyperbola $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\text{is}y=mx\pm \sqrt{b^2-a^2{m}^2}$
Thus equation of tangent to $9y^2-4x^2=1$ is
$y=mx\pm \sqrt{\frac{1}{9}-\frac{m^2}{4}}$
$\because$ Tangent is parallel to $3y=x+7$
$\therefore m=\frac{1}{3}$

Hence tangent will be
$y=\frac{x}{3}\pm \sqrt{\frac{1}{9}-\frac{1}{9\left(4\right)}}$
$\therefore 2x-6y=\pm \sqrt{3}$