If $P$ is a point on $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with foci $S$ and $S^{\prime}$,then the maximum area of $\triangle S P S^{\prime}$ is

If $F_1$ and $F_2$ are the foci of the ellipse $16 x^2+25 y^2=400$ and $P$ is any point on it,then the value of the product $P F_1 \cdot P F_2$ lies in the interval

The equation $2x^2 + 3y^2 = 30$ represents

The tangent to the ellipse $9x^2 + 16y^2 = 288$ making equal intercepts on the coordinate axes intersects the $X$-axis and the $Y$-axis at points $A$ and $B$ respectively. Then,the area of $\triangle OAB$ (where $O$ is the origin) is:

Find the equations of the tangents to the ellipse $3x^{2} + 4y^{2} = 12$ which are perpendicular to the line $y + 2x = 4$.

$A$ vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $PQR$,$\Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$,then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$