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Consider a tangent to the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$ at any point. The locus of the midpoint of the portion intercepted between the axes is
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Verified Answer
The correct answers are:
$\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$
Hint:
Tangent at $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$
$\frac{x x_{1}}{2}+\frac{y y_{1}}{1}=1$
Let mid point of intercept be $\mathrm{P}(\mathrm{h}, \mathrm{k})$
$\mathrm{h}=\frac{1}{\mathrm{x}_{1}}, \mathrm{k}=\frac{1}{2 \mathrm{y}_{1}}$ or $\mathrm{x}_{1}=\frac{1}{\mathrm{~h}}, \mathrm{y}_{1}=\frac{1}{2 \mathrm{k}}$
locus is $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$
Tangent at $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$
$\frac{x x_{1}}{2}+\frac{y y_{1}}{1}=1$
Let mid point of intercept be $\mathrm{P}(\mathrm{h}, \mathrm{k})$
$\mathrm{h}=\frac{1}{\mathrm{x}_{1}}, \mathrm{k}=\frac{1}{2 \mathrm{y}_{1}}$ or $\mathrm{x}_{1}=\frac{1}{\mathrm{~h}}, \mathrm{y}_{1}=\frac{1}{2 \mathrm{k}}$
locus is $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$
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