Question

The sum of $x-$intercept and $y-$intercept of the common tangent to the parabola $y^2=16x$ and $x^2=128y$ is

• A. $-8$
• B. $-12$
• C. $-16$
• D. $-24$

Answer 1

The correct option is D $-24$

Tangents to the given parabola's are given by,

$y=mx+\frac{4}{m}$ & $x=\frac{y}{m}+32m$

$mx-y+\frac{4}{m}=0$ and $\frac{y}{m}-x+32m=0$

$\because$ Both the lines are same,

$\Rightarrow \frac{m}{-1}=\frac{-1}{\frac{1}{m}}=\frac{\frac{4}{m}}{32m}$

$\Rightarrow -m=\frac{4}{32m^2}$

$\Rightarrow m^3=-\frac{1}{8}\Rightarrow m=-\frac{1}{2}$

Thus, required tangent is,

$x+2y+16=0$

$\therefore$ $x-$intercept $=-16$ and $y-$intercept $=-8$