Minimum distance between two points $P$ and $Q$ on the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{4} = 1$ , if difference between eccentric angles of $P$ and $Q$ is $\frac{{3\pi }}{2}$ , is
$2\sqrt 2 $
$2\sqrt 5 $
$\sqrt {29} $
$\sqrt {62} $
The line $x\cos \alpha + y\sin \alpha = p$ will be a tangent to the conic $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, if
The foci of $16{x^2} + 25{y^2} = 400$ are
The equation of the ellipse whose latus rectum is $8$ and whose eccentricity is $\frac{1}{{\sqrt 2 }}$, referred to the principal axes of coordinates, is
Consider an ellipse with foci at $(5,15)$ and $(21,15)$. If the $X$-axis is a tangent to the ellipse, then the length of its major axis equals
From the point$ C(0,\lambda )$ two tangents are drawn to ellipse $x^2\ +\ 2y^2\ = 4$ cutting major axis at $A$ and $B$. If area of $\Delta$ $ABC$ is minimum, then value of $\lambda$ is-