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If $\frac{\mathrm{x}}{\mathrm{ma}}+\frac{\mathrm{y}}{\mathrm{nb}}=1$ touches the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ then
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The correct answer is:
$\mathrm{m}^{2}=\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}$ or $\mathrm{n}^{2}=\frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}$
The line is $y=-\frac{\text { nb }}{\text { ma }} x-n b$. It will touch the ellipse if $(-n b)^{2}=a^{2}\left(-\frac{n b}{m a}\right)^{2}+b^{2}$
$\left[\mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}\right]$
$\Rightarrow n^{2}=\frac{n^{2}}{m^{2}}+1$
$\Rightarrow \mathrm{m}^{2}=\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}$ or $\mathrm{n}^{2}=\frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}$
$\left[\mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}\right]$
$\Rightarrow n^{2}=\frac{n^{2}}{m^{2}}+1$
$\Rightarrow \mathrm{m}^{2}=\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}$ or $\mathrm{n}^{2}=\frac{\mathrm{m}^{2}}{\mathrm{~m}^{2}-1}$
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