Find the equation of the ellipse whose center is at $(2, -3)$,focus is at $(3, -3)$,and one vertex is at $(4, -3)$.

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

If $P(\theta)$ and $Q\left(\frac{\pi}{2}+\theta\right)$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the locus of the midpoint of $PQ$ is $\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1$,then $\frac{a+b}{\alpha+\beta}=$

The equation to the locus of the middle point of the portion of the tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ included between the coordinate axes is:

Consider a tangent to the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$ at any point. The locus of the midpoint of the portion intercepted between the axes is