The equation of normal at the point (0, 3) of | vedclass

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(d) For $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,$ equation of normal at point $({x_1},{y_1})$,

==> $\frac{{(x - {x_1}){a^2}}}{{{x

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$y$ - axis

Vedclass · Answer

(d) For $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,$ equation of normal at point $({x_1},{y_1})$,

==> $\frac{{(x - {x_1}){a^2}}}{{{x_1}}} = \frac{{(y - {y_1}){b^2}}}{{{y_1}}}$;

$({x_1},{y_1}) \equiv (0,3),\,\,{a^2} = 5,\,\,{b^2} = 9$

$ \Rightarrow \frac{{(x - 0)}}{0}\,\,\,5 = \frac{{(y - 3).9}}{3}$ or $x = 0$ $i.e.$, $y$ - axis.

The equation of normal at the point $(0, 3)$ of the ellipse $9{x^2} + 5{y^2} = 45$ is

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