Question

Equation of common tangent of $y=x^2,y=-x^2+4x-4$ is

• A. $y=4(x-1)$
• B. $y=0$
• C. $y=-4(x-1)$
• D. $y=-30x-50$

Answer 1

Answer: (A), (B)

$y=0$

The equation of tangent to $y=x^2$, be $y=mx-\frac{m^2}{4}$. Putting in $y=-x^2+4x-4$, we should only get one value of x i.e. Discriminant must be zero.
$\therefore mx-\frac{m^2}{4}=-x^2+4x-4$
$\Rightarrow x^2+x(m-4)+4-\frac{m^2}{4}=0$
$D=0$
Now, $(m-4{)}^2-(16-m^2)=0$
$\Rightarrow 2m(m-4)=0\Rightarrow m=0,4$
$\therefore y=0$ and $y=4(x-1)$ are the required tangents.
Hence, (a) and (b) are correct answers.