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If tangents are drawn to the ellipse $x^{2}+2 y^{2}=2$ at all
points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :
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points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :
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Verified Answer
The correct answer is:
$\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$
Given the equation of ellipse,
$\frac{x^{2}}{(\sqrt{2})^{2}}+y^{2}=1$
$\frac{\sqrt{2} \cos \theta x}{2}+y \sin \theta=1$
$P\left(\frac{\sqrt{2}}{\cos \theta}, 0\right)$ and $Q\left(0, \frac{1}{\sin \theta}\right)$
Let mid point be $(h, k)$
$\Rightarrow h=\frac{1}{\sqrt{2} \cos \theta}, k=\frac{1}{2 \sin \theta}$
As $\cos ^{2} \theta+\sin ^{2} \theta=1$
$\therefore \quad \frac{1}{2 h^{2}}+\frac{1}{4 k^{2}}=1$
Locus is $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$
$\frac{x^{2}}{(\sqrt{2})^{2}}+y^{2}=1$
$\frac{\sqrt{2} \cos \theta x}{2}+y \sin \theta=1$
$P\left(\frac{\sqrt{2}}{\cos \theta}, 0\right)$ and $Q\left(0, \frac{1}{\sin \theta}\right)$
Let mid point be $(h, k)$
$\Rightarrow h=\frac{1}{\sqrt{2} \cos \theta}, k=\frac{1}{2 \sin \theta}$
As $\cos ^{2} \theta+\sin ^{2} \theta=1$
$\therefore \quad \frac{1}{2 h^{2}}+\frac{1}{4 k^{2}}=1$
Locus is $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$
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