In an ellipse $9x^2 + 5y^2 = 45$,the distance between the foci is

The extremities of the latera recta of the ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ having a given major axis $2a$ lie on:

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 $(R)$: 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:

Find the equation of the ellipse whose foci are $(\pm 2, 0)$ and eccentricity is $1/2$.

If the centre,one of the foci,and the semi-major axis of an ellipse are $(0, 0)$,$(0, 3)$,and $5$ respectively,then its equation is:

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