Question

The equation of the tangents to $4x^2-9y^2=36$ which are perpendicular to the straight line $2y+5x=10$ are

• A. $5(y-3)=2(x-\sqrt{\frac{117}{4}})$
• B. $5(y-2)=2(x-\sqrt{-18})$
• C. $5(y+2)=2(x-\sqrt{-18})$
• D. none of these

Answer 1

The correct option is D none of these

Given, $2y+5x=10$
$2y^{\prime }+5=0$
$y^{\prime }=\frac{-5}{2}$.
Hence the required slope is $\frac{2}{5}$.
Now
$4x^2-9y^2=36$.
$4x^2-36=9y^2$
$y=\frac{1}{3}.\sqrt{4x^2-36}$.
$y^{\prime }=\frac{8x}{6\sqrt{4x^2-36}}$ $=\frac{2}{5}$
$40x=12\sqrt{4x^2-36}$
$10x=3\sqrt{4x^2-36}$
$100x^2=36x^2-324$
$64x^2=-324$
$x^2=\frac{-81}{16}$.
Hence no real values of x.
Hence answer is none of these.