Question

Find the point of contact of tangents to the parabola $y^2=16x$ which are parallel and perpendicular to the line $2x-y+5=0.$

• A. $(16,-16)$
• B. $(16,16)$
• C. $(0,16)$
• D. $(0,-16)$

Answer 1

The correct option is A $(16,-16)$

We know that $y=mx+\frac{a}{m}$ is a tangent to the parabola $y^2=4ax$

And the point of contact is $(\frac{a}{m^2},\frac{2a}{m}).$
Here $y^2=16x$ is the parabola

$\therefore 4a=16$ or $a=4.$
Slope of the line $2x-y+5=0$ is $2.$
Any tangent parallel to it will have its slope $2$ and $\perp$ar to it will have slope $-\frac{1}{2}.$
Hence, putting $a=4,m=2$ and $-\frac{1}{2}$ the tangents are

$2x-y+2=0$ at $(1,4)$ and $x+2y+16=0$ at $(16,-16)$
Ans: B