A point on the ellipse, $4x^2 + 9y^2 = 36$, where the normal is parallel to the line, $4x -2y-5 = 0$ , is
$\left( {\frac{9}{5},\frac{8}{5}} \right)$
$\left( {\frac{8}{5},-\frac{9}{5}} \right)$
$\left( {-\frac{9}{5},\frac{8}{5}} \right)$
$\left( {\frac{8}{5},\frac{9}{5}} \right)$
The pole of the straight line $x + 4y = 4$ with respect to ellipse ${x^2} + 4{y^2} = 4$ is
Statement $-1$ : If two tangents are drawn to an ellipse from a single point and if they are perpendicular to each other, then locus of that point is always a circle
Statement $-2$ : For an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , locus of that point from which two perpendicular tangents are drawn, is $x^2 + y^2 = (a + b)^2$ .
If $m$ is the slope of a common tangent to the curves $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ and $x^{2}+y^{2}=12$, then $12\; m ^{2}$ is equal to
The length of the latus rectum of the ellipse $9{x^2} + 4{y^2} = 1$, is
For $0 < \theta < \frac{\pi}{2}$, four tangents are drawn at the four points $(\pm 3 \cos \theta, \pm 2 \sin \theta)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If $A(\theta)$ denotes the area of the quadrilateral formed by these four tangents, the minimum value of $A(\theta)$ is