JEE Advanced · 2006 · 09 Apr 2006 · Parabola

For the parabola $y=x^2$, the equation of a tangent with slope $m$ is $y=mx-\frac{m^2}{4}$.

For the parabola $y=-(x-2{)}^2$, which can be written as $(x-2{)}^2=-y$, the equation of a tangent with slope $m$ is $y=m(x-2)+\frac{m^2}{4}$.

For these to be the same line (common tangent), we need:







So $m=0$ or $m=4$.

For $m=0$: $y=0$

For $m=4$: $y=4x-\frac{16}{4}=4x-4=4(x-1)$

Therefore, the common tangents are $y=0$ and $y=4(x-1)$.