Question

The equation of a common tangent to the parabolas $y = x ^2$ and $y =-( x -2)^2$ is

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

Answer 1

Answer: (B)

Equation of tangent of $y=x^2$ be
$ t x=y+a t^2 ......$(1)
$ y=t x-\frac{t^2}{4}$
Solve with $y =-( x -2)^2$
$t x-\frac{t^2}{4}=-(x-2)^2 $
$x^2+x(t-4)-\frac{t^2}{4}+4=0$
$D=0$
$ (t-4)^2-4 \cdot\left(4-\frac{t^2}{4}\right)=0$
$t^2-4 t=0 $
$ t=0 $ or $ t=4$
From eq. (1), required common tangent is $y=4(x-1)$