Mathematics - Parabola Question with Solution | TestHub

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