Question

Number of common tangents of $y=x^2$ and $y=-x^2+4x-4$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

We have, equation of parabola $y=x^2$
Let $P\left(\alpha ,{\alpha }^2\right)$ is a point on the parabola,
$\therefore y-{\alpha }^2=2\alpha (x-\alpha )$
$\because \left[\because \frac{dy}{dx}=2x\Rightarrow \frac{dy}{d{x}_{\left(\alpha ,a^2\right)}}=2\alpha \right]$
$\Rightarrow y=2\alpha x-{\alpha }^2$
Also, given $y=-x^2+4x-4$
$\therefore -x^2+4x-4=20x-{\alpha }^2$
$\Rightarrow x^2+2x(\alpha -2)+\left(4-{\alpha }^2\right)=0$
Discriminant $=0$
$4(\alpha -2{)}^2-4\left(4-{\alpha }^2\right)=0$
$\Rightarrow (\alpha -2{)}^2-\left(4-{\alpha }^2\right)=0$
$\Rightarrow {\alpha }^2-4\alpha +4-4+{\alpha }^2=0$
$\Rightarrow {\alpha }^2-2\alpha =0$
$\Rightarrow \alpha =0,\alpha =2$