If $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord of an ellipse,then the eccentricity of the ellipse is

Assertion $(A)$: The length of the latus rectum of an ellipse is $4$. The focus and its corresponding directrix are respectively $(1, -2)$ and $3x + 4y - 15 = 0$. Then its eccentricity is $\frac{1}{2}$.
Reason $(R)$: The length of the perpendicular drawn from the focus of an ellipse to its corresponding directrix is $\frac{a(1 - e^2)}{e}$.
Which one of the following is correct?

The number of values of $c$ such that the line $y = cx + c$,where $c \in R$,touches the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ is:

The point $(1,3)$ with respect to the ellipse $4x^2+9y^2-16x-54y+61=0$ lies

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is

The locus of a point such that the sum of its distances from the points $(0, 2)$ and $(0, -2)$ is $6$ is: