The equation to the locus of the middle point | vedclass

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Question

Let any tangent of ellipse is $\frac{x \cos 	heta}{4}+\frac{y \sin 	heta}{3}=1$

Let it meets axes at $A\left(\frac{4}{\cos 	heta}, 0ight) \&am

Vedclass · Accepted Answer

$9x^2 + 16y^2 = 4 x^2y^2$

Vedclass · Answer

Let any tangent of ellipse is $\frac{x \cos 	heta}{4}+\frac{y \sin 	heta}{3}=1$

Let it meets axes at $A\left(\frac{4}{\cos 	heta}, 0ight) \& B\left(0, \frac{3}{\sin 	heta}ight)$

Let midpoint of $A B$ is $(h, k)$ then

$2 h=\frac{4}{\cos 	heta}, 2 k=\frac{3}{\sin 	heta}$

$\because \cos ^{2} 	heta+\sin ^{2} 	heta=1$

$	herefore \frac{16}{4 h^{2}}+\frac{9}{4 k^{2}}=1$

$\Rightarrow 16 k^{2}+9 h^{2}=4 h^{2} k^{2}$

Hence, locus is $16 y^{2}+9 x^{2}=4 x^{2} y^{2}$

The equation to the locus of the middle point of the portion of the tangent to the ellipse $\frac{{{x^2}}}{{16}}$$+$ $\frac{{{y^2}}}{9}$ $= 1$ included between the co-ordinate axes is the curve :

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