Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6, 0)$,foci $(\pm 4, 0)$.

$A$ rod $AB$ of length $15 \ cm$ rests between two coordinate axes such that the end point $A$ lies on the $x-$axis and the end point $B$ lies on the $y-$axis. $A$ point $P(x, y)$ is taken on the rod such that $AP = 6 \ cm$. Show that the locus of $P$ is an ellipse.

If the line $\alpha x+4y=\sqrt{7}$, where $\alpha \in R$, touches the ellipse $3x^{2}+4y^{2}=1$ at the point $P$ in the first quadrant, then one of the focal distances of $P$ is:

If the length and breadth of a rectangle of maximum area that can be inscribed in an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $8 \sqrt{2}$ and $4 \sqrt{2}$ respectively,then the eccentricity of that ellipse is

If the equations $x = 1 + 2 \cos \theta$ and $y = 2 + \sin \theta$ for $0 \leq \theta < 2 \pi$ represent an ellipse,then the point of intersection of the normal drawn at $P(\theta = \pi/4)$ to this ellipse and its major axis is:

The distance between the foci of an ellipse is $16$ and its eccentricity is $\frac{1}{2}$. The length of the major axis of the ellipse is: