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Find the normals to the ellipse $x^{2} / 9+y^{2} / 4=1$ which are farthest from its centre.
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Solution:

Given ellipse is, $ \frac{x^{2}}{9}+\frac{y^{2}}{4}=1........(1) \\ $.

From the symmetry in all the four quadrants,

let, $ P(3 \cos \phi, 2 \sin \phi), 0\lt\phi\lt\pi / 2 \\ $.

Equation of normal at $P(\phi)$ is,

$$ 3 \sec \phi x-2 \operatorname{cosec} \phi y=5 \\ $$

Distance of this normal from centre $(0,0) \\$ $$ =\mathrm{p}=\frac{5}{\sqrt{9 \sec ^{2} \phi+4 \operatorname{cosec}^{2} \phi}} \ $$

Obviously, p is maximum. When,

$Z=9 \sec ^{2} \phi+4 \operatorname{cosec}^{2} \phi $ is minimum.

Now $ \mathrm{dz} / \mathrm{d} \phi=18 \sec ^{2} \phi \tan \phi-8 \operatorname{cosec}^{2} \phi \cot \phi \\ $.

$ \mathrm{dz} / \mathrm{d} \phi=0$ gives $\tan ^{4} \phi=4 / 9 \quad$ i.e. $\tan \phi=\sqrt{(2 / 3)} \\ $.

Clearly for this value of, $ \phi, \mathrm{d}^{2} z / \mathrm{d} \phi^{2}\gt0 \\ $

So z is minimum when,

$ \tan \phi=\sqrt{2 / 3} \\ $

$\therefore$ Equation of required normals,

$$ \begin{aligned} \sqrt{3} 3 x-\sqrt{2} y &=\pm \sqrt{5} \quad \text { and } \quad \sqrt{3} x+\sqrt{2} y=\pm \sqrt{5}, \\\\ \text { Taking } \quad \phi &=\tan ^{-1} \sqrt{2 / 3}, \pi-\tan ^{-1} \sqrt{2 / 3},-\tan ^{-1} \sqrt{2 / 3}, \\\\ \pi &+\tan ^{-1} \sqrt{2 / 3} \end{aligned} \\ $$

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