Conics - How to Prove

Question

Not really sure how to approach part (iii) I have proved parts (i) and (ii), I'm assuming I have to use those answers. Any help would be greatly appreciated

Answer 1

$$x+p^2y=2cp$$ is the equation given to us, at P. Now where will it intersect at $x$-axis? At $y=0$.

$x=2cp$ and the coordinate according to this equation is , $(2cp,0)$.

But according to the condition given by locus it should be $(cq,0)$.

Hence $2cp=cq$ $\implies$ $2p=q$. This is our required condition for the locus. Putting this condition in $T(h,k)$ the point you have already found gives.

$$\left(\frac{4cp^2}{3p},\frac{2c}{3p}\right)\implies \left(\frac{4cp}{3},\frac{2c}{3p}\right)$$

Therefore locus is $$h=\frac{4cp}{3}$$ and $$k=\frac{2c}{3p}$$

Multiplying $h$,$k$ to eliminate the variable $p$.

$$hk=\frac{8}{9}c^2$$

Or the required locus is

$$xy=\frac{8}{9}c^2$$

Which is yet again a rectangular hyperbola rotated by $\frac{\pi}{4}$ and rectangular hyperbola always have eccentricity $\sqrt{2}$