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 $x^2 + y^2 = (a + b)^2$ .
Statement $-1$ is true, statement $-2$ is true but statement $-1$ is not the correct explanation for statement $-2$
Statement $-1$ is true, statement $-2$ is false
Statement $-1$ is false, statement $-2$ is true
Both statements are true, and statement $-1$ is the true explanation of statement $-2$
For an ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ with vertices $A$ and $ A', $ tangent drawn at the point $P$ in the first quadrant meets the $y-$axis in $Q $ and the chord $ A'P$ meets the $y-$axis in $M.$ If $ 'O' $ is the origin then $OQ^2 - MQ^2$ equals to
Eccentricity of the ellipse $9{x^2} + 25{y^2} = 225$ is
A rod of length $12 \,cm$ moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point $P$ on the rod, which is $3\, cm$ from the end in contact with the $x-$ axis.
If the line $x\cos \alpha + y\sin \alpha = p$ be normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then
Find the equation for the ellipse that satisfies the given conditions : Vertices $(\pm 5,\,0),$ foci $(\pm 4,\,0)$