Question

Equation of the tangent to $y^2=16x$ which perpendicular to $2x-y+5=0$ is

• A. $x+2y+16=0$
• B. $x+2y-17=0$
• C. $x+2y-19=0$
• D. $x+2y-18=0$

Answer 1

The correct option is A $x+2y+16=0$

$y^2=16x$

Differentiating above equation w.r.t. $x$, we have

$2y\frac{dy}{dx}=16\therefore \frac{dy}{dx}=\frac{8}{y}$

According to the question-

$\frac{dy}{dx}=\frac{-1}{2}$

$\therefore \frac{8}{y}=\frac{-1}{2}\Rightarrow y=-16$

$\therefore x=16$

$\therefore$ Equation of tangent $\Rightarrow (y+16)=\frac{-1}{2}(x-16)$

$\Rightarrow 2y+32=-x+16$

$\Rightarrow x+2y+16=0$