Question

If $CF$ is perpendicular from the centre $C$ of the ellipse $\frac{x^2}{49}+\frac{y^2}{25}=1$ on the tangent at any point $P$, and $G$ is the point where th e normal at $P$ meets the minor axis, then $(CF\cdot PG{)}^2$ is equal to___

Answer 1

Equation of the tangent at $P(7\cos \theta ,5\sin \theta )$ on the ellipse is $\frac{x}{7}\cos \theta +\frac{y}{5}\sin \theta =1\text{then}(CF{)}^2=\frac{7^2\times 5^2}{5^{2}co{s}^2\theta +7^{2}si{n}^2\theta }=\frac{25\times 49}{25{\cos }^2\theta +49{\sin }^2\theta }$
Equa tion of the normal at P is
$\frac{7x}{\cos \theta }-\frac{5y}{\sin \theta }=7^2-5^2$
Coordintes of $G$ are $(0,\frac{-24\sin \theta }{5})$
$(PG{)}^2=(7\cos \theta {)}^2+{(5\sin \theta +\frac{24\sin \theta }{5})}^2=\frac{49}{25}(25{\cos }^2\theta +49{\sin }^2\theta )So(CF\cdot PG{)}^2=(49{)}^2=2401$