Question

If the normal at $(1,2)$ on the parabola $y^2=4x$ meets the parabola again at the point $\left(t^2,2t\right)$ then $t$=

Answer 1

$y^2=4x$

Differentiating with respect to $x$,

$2y\frac{dy}{dx}=4$

$\frac{dy}{dx}=\frac{2}{y}$

Slope of normal $=-1\left(\frac{dy}{dx}\right)$

at $(1,2)$ slope of normal $=-1\left(\frac{2}{2}\right)=-1$

So, equation of normal, $(y-2)=-(x-1)x+y=3⟶\left(1\right)$

$(t^2,2t)$ is a point on $\left(1\right)$

$⟹t^2+2t=3$

$⟹t^2+2t-3=0$

$⟹t^2+3t-t-3=0$

$⟹(t+3)(t-1)=0$

$t=1,-3$

; t=3