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

In the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the equation of the diameter conjugate to the diameter $y = \frac{b}{a}x$ is:

Let $P$ be a point on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and let the perpendicular drawn through $P$ to the major axis meet its auxiliary circle at $Q$. If the normals drawn at $P$ and $Q$ to the ellipse and the auxiliary circle respectively meet in $R$,then the equation of the locus of $R$ is

$A$ tangent drawn at a point on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ cuts the $X$-axis at point $A$. If $A^{\prime}$ is the image of $A$ with respect to the line $y=x$,then the circle with $AA^{\prime}$ as its diameter passes through the fixed point:

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval:

If $S$ is the focus of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ lying on the positive $X$-axis and $P(\theta)$ is a point on the ellipse such that $SP=1$,then $\cos \theta=$