Question

A normal is drawn to a parabola $y^2=4ax$ at any point other than the vertex and it cuts the parabola again at a point whose distance from the vertex is not less than $\lambda \sqrt{6}$a, then the value of $\lambda$ is?

Answer 1

Parabola: $y^2=4ax..............\left(1\right)$

Let Normal at points other than vertex is drawn at,

$A(a{t}^2,2at)$ then it will the parabola again in

$B\left(a\right({-t-\frac{2}{t})}^2,2a(-t-\frac{2}{t}))$

Given distance is greater than or equal to $\lambda \sqrt{6}a$

So, $BV=\sqrt{[a(\frac{t^2+2}{t}{)}^2{]}^2+[2a\left(\frac{{-t}^2-2}{t}\right){]}^2}\ge \lambda \sqrt{6}a$

$=>\frac{a}{t}[(t^4+4+4t^2{)}^2+4(t^4+4+4t^2){]}^{\frac{1}{2}}\ge \lambda \sqrt{6}a$

$=>\frac{a}{t}[{\left({t^2+2)}^2\right(t^4+4+4t^2+4\left)\right]}^{\frac{1}{2}}\ge \lambda \sqrt{6}a$

$=>(t^2+2)[t^4+{4t}^2+8{]}^{\frac{1}{2}}\ge \lambda \sqrt{6}a$

Now, $\lambda \le \frac{1}{\sqrt{6}}(t^4+4t^2+8{)}^{\frac{1}{2}}.(t+\frac{2}{t})$