The line lx + my + n = 0is a normal to the el | vedclass

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(b) The equation of any normal to $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is

$ax\sec 	heta - $$by {m{cosec}}	heta = {a^2} - {b^

Vedclass · Accepted Answer

$\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$

Vedclass · Answer

(b) The equation of any normal to $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is

$ax\sec 	heta - $$by {m{cosec}}	heta = {a^2} - {b^2}$ .....(i)

The straight line $lx + my + n = 0$.....(ii)

will be a normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
If (i) and (ii) represent the same line

then, $\frac{{a\sec 	heta }}{l} = \frac{{b\,{m{cosec}}	heta }}{{ - m}} = \frac{{{a^2} - {b^2}}}{{ - n}}$

$ \Rightarrow \cos 	heta = \frac{{ - an}}{{l({a^2} - {b^2})}}$ and $\sin 	heta = \frac{{bn}}{{m({a^2} - {b^2})}}$

${\cos ^2}	heta + {\sin ^2}	heta = 1$

$\frac{{{a^2}{n^2}}}{{{l^2}{{({a^2} - {b^2})}^2}}} + \frac{{{b^2}{n^2}}}{{m{{({a^2} - {b^2})}^2}}} = 1$

==> $\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$.

The line $lx + my + n = 0$is a normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, if

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