Question

A normal is drawn to a parabola $y^2=4ax$ at any point other than the vertex . Prove that it cuts the parabola again at a point whose distance from the vertex is not less than $4\sqrt{6}a$

Answer 1

Any general point on $Y^2=4ax$

Can be written on $(a{t}^2,2at)$

When a normal meets the parabola again, then

$t^1=(t+\frac{2}{t})$

$\therefore$ Co-ordinate $ax=\left\{a\right(t+\frac{2}{t}{)}^2,2a(t+\frac{2}{t})\}$

vertex of parabola is $(0,0)$

$\therefore$ distance of point from vertex

$\sqrt{\left(a\right(t+\frac{2}{t}{)}^2{)}^2+\left(2a\right(t+\frac{2}{t}{)}^2)}$

$=a(t+\frac{2}{t})\sqrt{t+(\frac{2}{t}{)}^2+4}$

$=a(t+\frac{2}{t})\sqrt{t^2+\frac{4}{t^2}+4+4}$

$=a(t+\frac{2}{t})$

minimum value of $t+\frac{2}{t}$

$\frac{t+\frac{2}{t}}{2}\ge \sqrt{t\times \frac{2}{t}}$ $\{AM\ge GM\}$

$(t+\frac{2}{t})\ge 2\sqrt{2}$

$\therefore$ min value of $(t+\frac{2}{t})=2\sqrt{2}$

minimum value of

$\ell =a\times 2\sqrt{2}\sqrt{(2\sqrt{2}{)}^2+4}$

$2\sqrt{2}a\times \sqrt{12}$

$=4\sqrt{6}a$

hence, proved.