If the normal at any point P on the ellipse c | vedclass

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Question

(a) Let at a point $({x_1},{y_1})$ normal will be

$\frac{{(x - {x_1}){a^2}}}{{{x_1}}} = \frac{{(y - {y_1}){b^2}}}{{{y_1}}}$

At $G,\,\,y = 0$

Vedclass · Accepted Answer

${a^2}{(CG)^2} + {b^2}{(Cg)^2} = {({a^2} - {b^2})^2}$

Vedclass · Answer

(a) Let at a point $({x_1},{y_1})$ normal will be

$\frac{{(x - {x_1}){a^2}}}{{{x_1}}} = \frac{{(y - {y_1}){b^2}}}{{{y_1}}}$

At $G,\,\,y = 0$

==> $x = CG = \frac{{{x_1}({a^2} - {b^2})}}{{{a^2}}}$

At $g,\,\,\,x = 0$

==> $y = Cg = \frac{{{y_1}({b^2} - {a^2})}}{{{b^2}}}$

$\frac{{x_1^2}}{{{a^2}}} + \frac{{y_1^2}}{{{b^2}}} = 1$

==> ${a^2}{(CG)^2} + {b^2}{(Cg)^2} = {({a^2} - {b^2})^2}.$

If the normal at any point $P$ on the ellipse cuts the major and minor axes in $G$ and $g$ respectively and $C$ be the centre of the ellipse, then

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