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

Locus of the foot of the perpendicular drawn from the centre upon any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, is

Let $\mathrm{E}$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3,-4)$, one focus at $(4,-4)$ and one vertex at $(5,-4) .$ If $m x-y=4, m\,>\,0$ is a tangent to the ellipse $\mathrm{E}$, then the value of $5 \mathrm{~m}^{2}$ is equal to $.....$

Minimum area of the triangle by any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with the coordinate axes is

If the variable line $y = kx + 2h$ is tangent to an ellipse $2x^2 + 3y^2 = 6$ , then locus of $P(h, k)$ is a conic $C$ whose eccentricity equals

If the lines $x -2y = 12$ is tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at the point $\left( {3,\frac{-9}{2}} \right)$, then the length of the latus rectum of the ellipse is