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

Question

Aicharya · Accepted Answer

Equation of tangent of parabola $y=x^2$ be
$t x=y+a t^2$ ...(i)
$y=t x-\frac{t^2}{4}$
Solve with $\mathrm{y}=-(\mathrm{x}-2)^2$
$t x-\frac{t^2}{4}=-(x-2)^2$
$x^2+x(t-4)-\frac{t^2}{4}+4=0$
Here, Discriminant $=0$.
$(t-4)^2-4 \cdot\left(4-\frac{t^2}{4}\right)=0 \Rightarrow t^2-4 t=0 \Rightarrow t=0$
or $\mathrm{t}=4$
Put value of $t$ in eq. (i), then $y=4(x-1)$.

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

Solution