$A$ point on the curve $x=3 \cos \theta, y=2 \sin \theta$ at which the tangent is perpendicular to the $X$-axis is

$S^{\prime}$ is the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1, (b < 5)$ lying on the negative $X$-axis and $P(\theta)$ is a point on this ellipse. If the distance between the foci of this ellipse is $8$ and $S^{\prime}P = 7$,then $\theta =$

Find the condition for the line $ax + by + c = 0$ to be a normal to an ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$.

The line $lx + my + n = 0$ is a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if

If the tangents drawn from a point $P$ to the ellipse $4x^2+9y^2-16x+54y+61=0$ are perpendicular,then the locus of $P$ is

Let a focus of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $S(4, 0)$ and its eccentricity be $\frac{4}{5}$. If the point $P(3, \alpha)$ lies on $E$ and $O$ is the origin, then the area of $\triangle POS$ is equal to: (in $/ 5$)