Question

Find the equation of the common tangent to the parabolas $y^2=2x$ and $x^2=16y.$

• A. $x+2y-1=0$
• B. $x+2y+2=0$
• C. $x+2y+1=0$
• D. $x+2y+4=0$

Answer 1

Answer: (B)

$x+2y+2=0$

Given equations of parabola are
$y^2=2x$ and $x^2=16y.$
For $y^2=2x$
$a=\frac{1}{2}$
Equation of tangent to $y^2=2x$ is
$y=mx+\frac{1}{2m}.$ ....(1)
Since, it is tangent to $x^2=16y,$
$x^2=16(mx+\frac{1}{2m})$
Now, $D=0$
$\Rightarrow 256m^2=4\times \frac{-8}{m}$
$\Rightarrow m^3=-\frac{1}{8}$
$m=-\frac{1}{2}$
Put the value of $m$ in eqn $\left(1\right)$, we get
$y=-\frac{1}{2}x-1$
$\therefore x+2y+2=0$ is the equation of common tangent.