Question

Find the equation of the tangent to the parabola $y^2=8x$, which is perpendicular to the line $x-2y+6=0$.

• A. $x+2y-1=0$
• B. $3x+y+3=0$
• C. $2x+y+1=0$
• D. None of these

Answer 1

The correct option is C $2x+y+1=0$

$y^2=8x=4\left(2\right)x$

So, here, $a=2$

Slope of the given line :

$x-2y+6=0$ is $2y=x+6$.

$y=\frac{x}{2}+3$

So slope of the given line is $\frac{1}{2}$.

Since the tangent is perpendicular to the line, the slope of the tangent is -2.

Equation of tangent whose slope is m,

$y=mx+\frac{a}{m}$

Here, $m=-2,a=2$.

Equation of tangent is $y=-2x+2/(-2)⟹2x+y+1=0$.