Let $P$ be any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. If $S_1$ and $S_2$ are its foci,then the maximum area of $\Delta PS_1S_2$ is (in square units):

Consider an ellipse,whose centre is at the origin and its major axis is along the $x-$ axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is $6$,then the area (in sq. units) of the quadrilateral inscribed in the ellipse,with the vertices as the vertices of the ellipse,is

Let $A(\alpha, 0)$ and $B(0, \beta)$ be the points on the line $5x + 7y = 50$. Let the point $P$ divide the line segment $AB$ internally in the ratio $7:3$. Let $3x - 25 = 0$ be a directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the corresponding focus be $S$. If from $S$,the perpendicular on the $x$-axis passes through $P$,then the length of the latus rectum of $E$ is equal to

Planet $M$ orbits around its sun,$S$,in an elliptical orbit with the sun at one of the foci. When $M$ is closest to $S$,it is $2$ units away. When $M$ is farthest from $S$,it is $18$ units away. Assuming $S$ is at the origin $(0, 0)$ and the other focus lies on the negative $y$-axis,find the equation of the elliptical orbit of planet $M$.

The coordinates of a point,in the parametric form,on the ellipse whose foci are $(-1, 0)$ and $(7, 0)$ and eccentricity $e = \frac{1}{2}$,are

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: