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 $\frac{a+b}{\alpha+\beta}=$

If $F_1$ and $F_2$ are the feet of the perpendiculars from the foci $S_1$ and $S_2$ of an ellipse $\frac{x^2}{5} + \frac{y^2}{3} = 1$ on the tangent at any point $P$ on the ellipse,then $(S_1 F_1) (S_2 F_2)$ is equal to

Find the eccentricity of an ellipse,if the length of its latus rectum is $4$ units and the distance between its vertex and the nearest focus is $3/2$ units.

If $x+2y+k=0, k>0$ is a tangent to the ellipse $2x^2+y^2=2$,then the equation of the normal to the given ellipse at $\left(\frac{1}{\sqrt{2}}, \frac{k}{3}\right)$ 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:

Let $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ be an ellipse. Ellipses $E_i$ are constructed such that their centres and eccentricities are the same as that of $E_1$,and the length of the minor axis of $E_i$ is the length of the major axis of $E_{i+1}$ $(i \geq 1)$. If $A_i$ is the area of the ellipse $E_i$,then $\frac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right)$ is equal to . . . . . . .