If $a x^2+b y^2=15$ is the equation of the ellipse for which distance between its foci is 2 and distance between its directrices is 5 , then $a+b=$

Assertion (A) The image of $\frac{x^2}{25}+\frac{y^2}{16}=1$ in the line $x+y=10$ is $\frac{(x-10)^2}{16}+\frac{(y-10)^2}{25}=1$

Reason ( $\mathbf{R}$ ) The image of a curve ' $C$ ' in a line $L$ is the locus of the image of every point of $C$ with respect to the line $L$. The correct option among the following is :

The equation of the normal to the curve $4 x^2+9 y^2=36$ at the point $P\left(\frac{7 \pi}{4}\right)$ is

Let $S \equiv \frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0, S \equiv \frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}-1=0$ be two intersecting ellipses. If $P(a \cos \theta, b \sin \theta)$ and $Q\left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right)$ are their points of intersection then $\frac{1}{2}\left(a^2 \beta^2+b^2 \alpha^2\right)=$