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If tangents are drawn to the ellipse $x^2+2 y^2=2$, then the locus of the mid-points of the intercepts made by those tangents between the coordinate axes is
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Verified Answer
The correct answer is:
$\frac{1}{2 x^2}+\frac{1}{4 y^2}=1$
Given ellipse,
$x^2+2 y^2=2 \Rightarrow \frac{x^2}{2}+\frac{y^2}{1}=1$
Point $P(\sqrt{2} \cos \theta, \sin \theta)$ lie in ellipse
Tangent at $P(\sqrt{2} \cos \theta, \sin \theta)$ on ellipse is,
$x \cos +\sqrt{2} \sin \theta y=\sqrt{2}$
intercept on line is, $A\left(\frac{\sqrt{2}}{\cos \theta}, 0\right)$ and $B\left(0, \frac{1}{\sin \theta}\right)$
Let $(h, k)$ is mid-point of the $A B$.
$\begin{aligned}
& \therefore \quad h=\frac{\sqrt{2}}{2 \cos \theta} \text { and } k=\frac{1}{2 \sin \theta} \\
& \Rightarrow \quad \cos \theta=\frac{1}{\sqrt{2} h} \text { and } \sin \theta=\frac{1}{2 k} \\
&
\end{aligned}$
Squaring and adding, we get
$\frac{1}{2 h^2}+\frac{1}{4 k^2}=1$
$\therefore$ Locus is, $\frac{1}{2 x^2}+\frac{1}{4 y^2}=1$
$x^2+2 y^2=2 \Rightarrow \frac{x^2}{2}+\frac{y^2}{1}=1$
Point $P(\sqrt{2} \cos \theta, \sin \theta)$ lie in ellipse
Tangent at $P(\sqrt{2} \cos \theta, \sin \theta)$ on ellipse is,
$x \cos +\sqrt{2} \sin \theta y=\sqrt{2}$
intercept on line is, $A\left(\frac{\sqrt{2}}{\cos \theta}, 0\right)$ and $B\left(0, \frac{1}{\sin \theta}\right)$
Let $(h, k)$ is mid-point of the $A B$.
$\begin{aligned}
& \therefore \quad h=\frac{\sqrt{2}}{2 \cos \theta} \text { and } k=\frac{1}{2 \sin \theta} \\
& \Rightarrow \quad \cos \theta=\frac{1}{\sqrt{2} h} \text { and } \sin \theta=\frac{1}{2 k} \\
&
\end{aligned}$
Squaring and adding, we get
$\frac{1}{2 h^2}+\frac{1}{4 k^2}=1$
$\therefore$ Locus is, $\frac{1}{2 x^2}+\frac{1}{4 y^2}=1$
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