$\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^{2}-b^{2}$
It meets axes at $Q\left(\frac{\left(a^{2}-b^{2}\right) \cos \theta}{a}, 0\right)$
and $R\left(0,-\frac{\left(a^{2}-b^{2}\right) \sin \theta}{b}\right)$
Let $T(h, k)$ is a midpoint of $Q R$ Then $2 h=\frac{\left(a^{2}-b^{2}\right) \cos \theta}{a}$
and $2 k=-\frac{\left(a^{2}-b^{2}\right) \sin \theta}{b}$
$\Rightarrow \cos ^{2} \theta+\sin ^{2} \theta=\frac{4 h^{2} a^{2}}{\left(a^{2}-b^{2}\right)^{2}}+\frac{4 k^{2} b^{2}}{\left(a^{2}-b^{2}\right)^{2}}=1$
$\Rightarrow$ Locus is $\frac{x^{2}}{\frac{\left(a^{2}-b^{2}\right)^{2}}{4 a^{2}}}+\frac{y^{2}}{\frac{\left(a^{2}-b^{2}\right)^{2}}{4 b^{2}}}=1$
which is an ellipse, having eccentricity $e^{\prime},$ given by $e^{r 2}=1-\frac{\frac{\left(a^{2}-b^{2}\right)^{2}}{4 a^{2}}}{\frac{\left(a^{2}-b^{2}\right)^{2}}{4 b^{2}}}=1-\frac{b^{2}}{a^{2}}=e^{2}$
$e^{\prime}=e$
Note : In Equation (ii), $\frac{\left(a^{2}-b^{2}\right)}{4 a^{2}}<\frac{\left(a^{2}-b^{2}\right)}{4 b^{2}}$. Hence, $x$ -axis is minor axis.
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