The foci of $16x^2 + 25y^2 = 400$ are

The parametric form of a point on the ellipse whose foci are $(-1, 0)$ and $(7, 0)$ and eccentricity is $1/2$ is:

The tangent and normal to the ellipse $3x^2 + 5y^2 = 32$ at the point $P(2, 2)$ meet the $x$-axis at $Q$ and $R$,respectively. Then the area (in sq. units) of the triangle $PQR$ is

If any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ cuts off intercepts of length $h$ and $k$ on the axes,then $\frac{a^2}{h^2} + \frac{b^2}{k^2} = $

The equation of the tangent to the curve $9x^{2} + 16y^{2} = 144$ which makes equal intercepts with the coordinate axes is:

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval: