A chord $PQ$ of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ subtends right angle at its  centre. The locus of the point of intersection of tangents drawn at $P$ and $Q$ is-

Let the line $y=m x$ and the ellipse $2 x^{2}+y^{2}=1$ intersect at a ponit $\mathrm{P}$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $\left(-\frac{1}{3 \sqrt{2}}, 0\right)$ and $(0, \beta),$ then $\beta$ is equal to

The point $(4, -3)$ with respect to the ellipse $4{x^2} + 5{y^2} = 1$

An ellipse having foci at $(3, 1)$ and $(1, 1) $ passes through the point $(1, 3),$ then its eccentricity is

Let $PQ$ be a focal chord of the parabola $y^{2}=4 x$ such that it subtends an angle of $\frac{\pi}{2}$ at the point $(3, 0)$. Let the line segment $PQ$ be also a focal chord of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$. If $e$ is the eccentricity of the ellipse $E$, then the value of $\frac{1}{e^{2}}$ is equal to

An ellipse has $OB$ as semi minor axis, $F$ and $F'$ its foci and the angle $FBF'$ is a right angle. Then the eccentricity of the ellipse is