# Question #00f1a

Apr 7, 2017

The given equation of the parabola

$y = 4 a {x}^{2}$

$\implies {x}^{2} = 4 \times \frac{1}{16 a} y$

Hence the parabola is symmetric about y-axis i,e its axis is Y-axis .

The co-ordinates of its vertex is $\implies \left(0 , 0\right)$.

The co-ordinates of its focus ( F)$\implies \left(0 , \frac{1}{16 a}\right)$

The equation of its directrix,$\implies y = - \frac{1}{16 a}$

Let the co-ordinates of any point P on it in parametric form be $\left(\frac{2 t}{16 a} , {t}^{2} / \left(16 a\right)\right)$,where $t$ is the parameter.

Now slope of the tangent at P

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\frac{2 t}{16 a} , {t}^{2} / \left(16 a\right)\right)} = 4 a \times 2 \times \frac{2 t}{16 a} = t$

So slope of the normal at P $= - \frac{1}{t}$

So equation of the normal at P

$y - {t}^{2} / \left(16 a\right) = - \frac{1}{t} \times \left(x - \frac{2 t}{16 a}\right)$

Putting ,$y = - \frac{1}{16 a}$ in the equation of the normal we get the X=coordinate of the point of intersection (Q) of the normal with the directrix

$- \frac{1}{16 a} - {t}^{2} / \left(16 a\right) = - \frac{1}{t} \times \left(x - \frac{2 t}{16 a}\right)$

$\implies - \frac{1}{16 a} - {t}^{2} / \left(16 a\right) = - \frac{x}{t} + \frac{2}{16 a}$

$\implies \frac{x}{t} = {t}^{2} / \left(16 a\right) + \frac{3}{16 a}$

$\implies x = {t}^{3} / \left(16 a\right) + \frac{3 t}{16 a}$

So co-ordinates of Q is $\left({t}^{3} / \left(16 a\right) + \frac{3 t}{16 a} , - \frac{1}{16 a}\right)$

So

$P {Q}^{2} = \left[{\left({t}^{3} / \left(16 a\right) + \frac{3 t}{16 a} - \frac{2 t}{16 a}\right)}^{2} + {\left({t}^{2} / \left(16 a\right) + \frac{1}{16 a}\right)}^{2}\right]$

$= \frac{1}{16 a} ^ 2 \left[{\left({t}^{3} + t\right)}^{2} + {\left({t}^{2} + 1\right)}^{2}\right]$

$= \frac{1}{16 a} ^ 2 \left[{t}^{2} {\left({t}^{2} + t\right)}^{2} + {\left({t}^{2} + 1\right)}^{2}\right]$

$= \frac{1}{16 a} ^ 2 \left[{\left({t}^{2} + t\right)}^{2} \left({t}^{2} + 1\right)\right]$

$= \frac{1}{16 a} ^ 2 \left[{\left({t}^{2} + t\right)}^{3}\right]$

No

$F {P}^{2} = {\left(\frac{2 t}{16 a}\right)}^{2} + {\left({t}^{2} / \left(16 a\right) - \frac{1}{16 a}\right)}^{2} = {\left({t}^{2} + 1\right)}^{2} / {\left(16 a\right)}^{2}$

$\implies F {P}^{3} = {\left({t}^{2} + 1\right)}^{3} / {\left(16 a\right)}^{3}$

Hence

$\frac{P {Q}^{2}}{F {P}^{3}} = 16 a$

$\implies P {Q}^{2} = 16 a F {P}^{3}$