Question

The product of the perpendicular from two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ on the tangent at any point on the ellipse is

• A. $8$
• B. $16$
• C. $12$
• D. $9$

Answer 1

The correct option is A $8$

$\frac{x^2}{9}+\frac{y^2}{25}=1$

$e=\sqrt{\frac{1-9}{25}}=\frac{4}{25}$

Let $(a,b)$ be on elipse and tangent is drawn at this points

$\frac{a^2}{9}+\frac{6^2}{25}=1$ ------(1)

Tangent equation $\to 25ax+69y-225=0\to T$

$\perp$ distance from $f$to $T=\frac{|366-225|}{\sqrt{(25a{)}^2+(96{)}^2}}$

$\perp$ distance from $f^{\prime }$ to $T=\frac{|366-225|}{\sqrt{(25a{)}^2+(96{)}^2}}$

product of above $2=\frac{|(366{)}^2|-(225{)}^2|}{625a^2+{516}^2}$

$a^2=9\frac{{96}^2}{25}$ $25a^2=225-{96}^2$

$625a^2=225\times 25-{2256}^2$

from (3) $\to \frac{|(366{)}^2-(225{)}^2|}{225\times 25-{1446}^2}=\frac{|(225{)}^2-(366{)}^2|}{225\times 25-{1446}^2}$

Taking $9$ common from numerate

$\frac{9|225\times 25-{1446}^2|}{225\times 25-{1446}^2}=9$