The line $y=x+1$ meets the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter,then $(3r)^{2}$ is equal to

Consider the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let $S(p, q)$ be a point in the first quadrant such that $\frac{p^2}{9}+\frac{q^2}{4}>1$. Two tangents are drawn from $S$ to the ellipse,of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive $x$-coordinate and $O$ be the center of the ellipse. If the area of the triangle $\triangle ORT$ is $\frac{3}{2}$,then which of the following options is correct?

$A$ 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:

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5, 0)$,foci $(\pm 4, 0)$.

If a circle $(x-1)^2+y^2=r^2$ touches the ellipse $x^2+4y^2=16$ internally,then $r=$