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. $-16$
• B. $-32$
• C. $-8$
• D. $-24$

Answer 1

The correct option is D $-24$

Tangent of slope $m$ to the parabola $y^2=16x$ is
$y=mx+\frac{4}{m}\dots \left(1\right)$

Equation $\left(1\right)$ is also a tangent to $x^2=128y$, so
$x^2=128(mx+\frac{4}{m})$ should have equal roots in $x$
i.e., for $m{x}^2-128m^{2}x-512=0$
$D=0\Rightarrow (128{)}^2{m}^4+4\times 512m=0\Rightarrow 128m(128m^3+16)=0\Rightarrow 128m^3+16=0(\because m\ne 0)\Rightarrow m=-\frac{1}{2}$

Therefore, the equation of common tangent is
$y=-\frac{x}{2}-8\Rightarrow \frac{x}{-16}+\frac{y}{-8}=1$

Hence, the sum of intercepts
$=-16-8=-24$