The area (in square units) of the quadrilateral formed by joining the foci of the two ellipses $\frac{x^2}{9}+\frac{y^2}{5}=1$ and $\frac{x^2}{5}+\frac{y^2}{9}=1$ is

The equation $\frac{x^2}{2-r}+\frac{y^2}{r-5}+1=0$ represents an ellipse if

Let the point $L$ lying in the first quadrant be one end of a latus rectum of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$. Let $P$ and $Q$ be the points where the normal drawn at $L$ to this given ellipse meets the major axis and the minor axis. Then the distance between $P$ and $Q$ is

Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3, 0)$,$a = 4$.

If $S$ and $S^{\prime}$ are the foci of an ellipse $\frac{x^2}{169}+\frac{y^2}{144}=1$ and the point $B$ lying on the positive $Y$-axis is one end of its minor axis,then the incenter of the triangle $SBS^{\prime}$ is

Let $P$ be a variable point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with foci $F_1$ and $F_2$. If $A$ is the area of the triangle $PF_1F_2$,then the maximum value of $A$ is