If $P(x, y)$,$F_1 = (3, 0)$,$F_2 = (-3, 0)$ and $16x^2 + 25y^2 = 400$,then $PF_1 + PF_2 = \dots$

On the ellipse $4x^{2} + 9y^{2} = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are:

An ellipse having the coordinate axes as its axes and its major axis along the $Y$-axis,passes through the point $(-3, 1)$ and has eccentricity $e = \sqrt{\frac{2}{5}}$. Then its equation is:

Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$. Then $n(S \cap T)$ is equal to $......$

From the point $C(0, \lambda)$,two tangents are drawn to the ellipse $x^2 + 2y^2 = 4$,cutting the major axis at $A$ and $B$. If the area of $\Delta ABC$ is minimum,then the value of $\lambda$ is-

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse is: