Question

Find the equations of the tangents to the parabola $y^2=16x$, which are parallel & perpendicular respectively to the line $2x-y+5=0$ ?

• A. $2x-y-2=0;x+2y+8=0$
• B. $2x-y+2=0;x+2y+8=0$
• C. $2x-y-2=0;x+2y+16=0$
• D. $2x-y+2=0;x+2y+16=0$

Answer 1

The correct option is D $2x-y+2=0;x+2y+16=0$

Equation of line parallel to line $2x-y+5=0$ is given by,

$y=2x+c$

Using condition of tangency, $c=\frac{a}{m}=\frac{4}{2}=2$
So, tangent parallel to the given line is $2x-y+2=0$
Now equation of line perpendicular to line $2x-y+5=0$ is given by

$y=-\frac{1}{2}x+c$

Using condition of tangency, $c=\frac{4}{-1/2}=-8$
So tangent perpendicular to the given line is, $x+2y+16=0$