Prove that the straight line $$lx + my = n$$ is a normal to the ellipse, if $$\dfrac {a^{2}}{l^{2}} + \dfrac {b^{2}}{m^{2}} = \dfrac {(a^{2} - b^{2})^{2}}{n^{2}}$$.
We have,
Equation of straight line is
$$lx+my=n\,\,......\,\,\left( 1 \right)$$
Normal to the ellipse
$$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\,\,\,\left( a>b \right)$$
We know that equation of line
$$y=mx+c$$
Equation of line normal to the ellipse be
$$y=mx\pm \dfrac{m\left( {{a}^{2}}-{{b}^{2}} \right)}{\sqrt{{{a}^{2}}+{{b}^{2}}{{m}^{2}}}}$$
On comparing that,
$$c=\pm \dfrac{m\left( {{a}^{2}}-{{b}^{2}} \right)}{\sqrt{{{a}^{2}}+{{b}^{2}}{{m}^{2}}}}\,\,......\,\,\left( 2 \right)$$
From (1),
$$ lx+my+n=0 $$
$$ \Rightarrow y=-\dfrac{l}{m}x-\dfrac{n}{m} $$
Now,
$$y=mx+c$$
On comparing that,
$$ c=\dfrac{-n}{m}\,\,......\,\,\left( 3 \right) $$
$$ slope\,=\dfrac{-l}{m} $$
From (2) and (3) to, and we get,
$$\pm \dfrac{m\left( {{a}^{2}}-{{b}^{2}} \right)}{\sqrt{{{a}^{2}}+{{b}^{2}}{{m}^{2}}}}=\dfrac{-n}{m}\,$$
On squaring both side and we get,
$$ \pm \dfrac{\left( \dfrac{-l}{m} \right)\left( {{a}^{2}}-{{b}^{2}} \right)}{\sqrt{{{a}^{2}}+{{b}^{2}}{{m}^{2}}}}=\dfrac{-n}{m}\, $$
$$ \dfrac{{{n}^{2}}}{{{m}^{2}}}=\dfrac{\dfrac{{{l}^{2}}}{{{m}^{2}}}{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}}{{{a}^{2}}+{{b}^{2}}\dfrac{{{l}^{2}}}{{{m}^{2}}}} $$
$$ \dfrac{{{a}^{2}}{{m}^{2}}+{{b}^{2}}{{l}^{2}}}{{{l}^{2}}{{m}^{2}}}=\dfrac{{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}}{{{n}^{2}}} $$
$$ \dfrac{{{a}^{2}}}{{{l}^{2}}}+\dfrac{{{b}^{2}}}{{{m}^{2}}}=\dfrac{{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}}{{{n}^{2}}} $$
Hence proved.
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