Question

If the normal at a point $P$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ of semi axes $a,b$ & centre $C$ cuts the major &minor axes at $G$ & $g$ , then

• A. $a^2.(CG{)}^2+b^2.(Cg{)}^2=(a^2-b^2{)}^2.$
• B. $CG=e^{2}CN,$ where $PN$ is the ordinate of $P.(N$ is foot of perpendicular from $P$ on its major axis.$)$
• C. $a^2.(CG{)}^2+b^2.(Cg{)}^2=(a^2-b^2).$
• D. $CG{e}^2=CN,$ where $PN$ is the ordinate of $P.(N$ is foot of perpendicular from $P$ on its major axis.$)$

Answer 1

Answer: (A), (B)

The correct options are
A $a^2.(CG{)}^2+b^2.(Cg{)}^2=(a^2-b^2{)}^2.$
B $CG=e^{2}CN,$ where $PN$ is the ordinate of $P.(N$ is foot of perpendicular from $P$ on its major axis.$)$

Equation of normal is $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$

So, $G=(\frac{x_1}{a^2}{a}^2-b^2,0)$

$g=(0,\frac{{-y}_1}{b^2}{a}^2-b^2)$

$a^2{CG}^2+b^2{Cg}^2=a^2\frac{{x_1}^2}{a^4}{(a^2-b^2)}^2+b^2\frac{{y_1}^2}{b^4}{(a^2-b^2)}^2$

$a^2{CG}^2+b^2{Cg}^2={(a^2-b^2)}^2(\frac{{x_1}^2}{a^2}+\frac{{y_1}^2}{b^2})$

$={(a^2-b^2)}^2$ as $(\frac{{x_1}^2}{a^2}+\frac{{y_1}^2}{b^2})=1$

$CG=\frac{{x_1}^2}{a^2}(a^2-b^2)=x_1\frac{a^2{e}^2}{a^2}=x_1{e}^2=e^{2}CN$