The eccentricity of the ellipse $(x-3)^2 + (y-4)^2 = \frac{y^2}{9} + 16$ 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:

The line $x \cos \alpha + y \sin \alpha = p$ will be a tangent to the conic $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if

Let the length of a latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $10$. If its eccentricity is the minimum value of the function $f(t) = t^2 + t + \frac{11}{12}$,$t \in R$,then $a^2 + b^2$ is 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:

The total number of tangents through the point $(3,5)$ that can be drawn to the ellipses $3x^2 + 5y^2 = 32$ and $25x^2 + 9y^2 = 450$ is