The length of the latus rectum of the ellipse $16x^2 + 25y^2 = 400$ is

For what values of $t$ does the line $x = at^2$ intersect the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at real points?

If the latus rectum of an ellipse subtends a right angle at the centre of that ellipse,then the eccentricity of that ellipse is

Consider the conic $e x^2 + \pi y^2 - 2 e^2 x - 2 \pi^2 y + e^3 + \pi^3 = \pi e$.
Suppose $P$ is any point on the conic and $S_1, S_2$ are the foci of the conic,then the maximum value of $(P S_1 + P S_2)$ is

An ellipse having foci at $(3, 3)$ and $(-4, 4)$ and passing through the origin has eccentricity equal to

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: