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

Let $P$ be an arbitrary point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $a > b > 0$. Suppose $F_1$ and $F_2$ are the foci of the ellipse. The locus of the centroid of the $\Delta P F_1 F_2$ as $P$ moves on the ellipse is

Let the foci and length of the latus rectum of an ellipse $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ be $( \pm 5,0)$ and $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{\mathrm{x}^2}{\mathrm{~b}^2}-\frac{\mathrm{y}^2}{\mathrm{a}^2 \mathrm{~b}^2}=1$ equals

The locus of point of intersection of two perpendicular tangent of the ellipse  $\frac{{{x^2}}}{{{9}}} + \frac{{{y^2}}}{{{4}}} = 1$ is :-

The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ is

The equation of the ellipse whose centre is $(2, -3)$, one of the foci is $(3, -3)$ and the corresponding vertex is $(4, -3)$ is