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

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(A) Let the point $P$ be $(x_1, y_1)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.The equation of the normal at $(x_1, y_1)$ is given by $\

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the point $P$ be $(x_1, y_1)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.The equation of the normal at $(x_1, y_1)$ is given by $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$.For the point $G$ on the major axis ($x$-axis),set $y = 0$:$\frac{a^2x}{x_1} = a^2 - b^2 \implies CG = x = \frac{x_1(a^2 - b^2)}{a^2}$.For the point $g$ on the minor axis ($y$-axis),set $x = 0$:$-\frac{b^2y}{y_1} = a^2 - b^2 \implies Cg = y = -\frac{y_1(a^2 - b^2)}{b^2}$.Now,calculate $a^2(CG)^2 + b^2(Cg)^2$:$a^2 \left[ \frac{x_1^2(a^2 - b^2)^2}{a^4} \right] + b^2 \left[ \frac{y_1^2(a^2 - b^2)^2}{b^4} \right] = (a^2 - b^2)^2 \left( \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} \right)$.Since $(x_1, y_1)$ lies on the ellipse,$\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1$.Therefore,$a^2(CG)^2 + b^2(Cg)^2 = (a^2 - b^2)^2$.

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

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