$A$ rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. Then its dimensions are:

The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

The area of the region bounded by the curve $x = 4 \cos \theta, y = 3 \sin \theta$ is . . . . . . sq. units. (in $\pi$)

Assertion $(A)$: If the tangent and normal to the ellipse $9x^2 + 16y^2 = 144$ at the point $P(\frac{\pi}{3})$ on it meet the major axis in $Q$ and $R$ respectively,then $QR = \frac{57}{8}$.
Reason $(R)$: If the tangent and normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $P(\theta)$ on it meet the major axis in $Q$ and $R$ respectively,then $QR = \left| \frac{a^2 \sin^2 \theta + b^2 \cos^2 \theta}{a \cos \theta} \right|$.

The focus of an ellipse is at the origin. The line $x = 4$ is a directrix and the eccentricity is $1/2$. Find the length of the major axis. (in $/3$)

Let $\frac{x^2}{f(a^2 + 7a + 3)} + \frac{y^2}{f(3a + 15)} = 1$ represent an ellipse with major axis along the y-axis, where $f$ is a strictly decreasing positive function on $R$. If the set of all possible values of $a$ is $R - [\alpha, \beta]$, then $\alpha^2 + \beta^2$ is equal to: