A tangent to the ellipse frac{x^2}{25}+frac{y | vedclass

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Question

Any tangent to the ellipse is

$\frac{x \cos 	heta}{5}+\frac{y \sin 	heta}{4}=1$

$	herefore $ $A = (5\sec 	heta ,0),B = (0,4\cos ec	heta )$

Vedclass · Accepted Answer

$\frac{25}{x^2}+\frac{16}{y^2}=4$

Vedclass · Answer

Any tangent to the ellipse is

$\frac{x \cos 	heta}{5}+\frac{y \sin 	heta}{4}=1$

$	herefore $ $A = (5\sec 	heta ,0),B = (0,4\cos ec	heta )$

Let $(\mathrm{h}, \mathrm{k})$ be the circumcentre of $\Delta \mathrm{A} \mathrm{OB},$ then

$2 \mathrm{h}=\frac{5}{\cos 	heta}$ and $2 \mathrm{k}=\frac{4}{\sin 	heta}$

$	herefore $ Locus of $(\mathrm{h}, \mathrm{k})$ is $\frac{25}{\mathrm{x}^{2}}+\frac{16}{\mathrm{y}^{2}}=4.$

A tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ intersect the co-ordinate axes at $A$ and $B,$ then locus of circumcentre of triangle $AOB$ (where $O$ is origin) is

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