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Q.
A tangent to the hypererbola $\frac{x^2}{4}-\frac{y^2}{1}=1$ meets ellipse $x^2+4 y^2=4$ at two distinct points. The locus of mid point of this chord is
Conic Sections
Solution:
Any tangent to given hypererbola, is
$\frac{x \cdot 2 \sec \theta}{4}-\frac{y \cdot \tan \theta}{1}=1$
Let $\left( x _1, y _1\right)$ be the middle point of the chord of ellipse.
$\therefore $ Its equation is $\frac{ xx _1}{4}+\frac{ yy _1}{1}=\frac{ x _1^2}{4}+\frac{ y _1^2}{1}$
As (1) and (2) are identical, so
$\frac{2 \sec \theta}{x_1}=\frac{-\tan \theta}{y_1}=\frac{1}{\frac{x_1^2}{4}+\frac{y_1^2}{1}} \Rightarrow \sec \theta=\frac{2 x_1}{x_1^2+4 y_1^2} \text { and } \tan \theta=\frac{-4 y_1}{x_1^2+4 y_1^2}$
As, $\sec ^2 \theta-\tan ^2 \theta=1$, so $\frac{4 x_1^2}{\left(x_1^2+4 y_1^2\right)^2}-\frac{16 y_1^2}{\left(x_1^2+4 y_1^2\right)^2}=1$
$\Rightarrow $ Locus is $\left(x^2+4 y^2\right)^2=4\left(x^2-4 y^2\right)$.