Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3, 0)$,$a = 4$.

The length of the latus rectum of the ellipse $\frac{x^2}{36} + \frac{y^2}{49} = 1$ is:

$A$ wall is inclined to the floor at an angle of $135^{\circ}$. $A$ ladder of length $l$ is resting on the wall. As the ladder slides down,its mid-point traces an arc of an ellipse. Then,the area of the ellipse is

$A$ tangent is drawn at $(3 \sqrt{3} \cos \theta, \sin \theta)$ $\left(0 < \theta < \frac{\pi}{2}\right)$ to the ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$. The value of $\theta$ for which the sum of the intercepts on the coordinate axes made by this tangent attains the minimum is

The equations of the tangents to the ellipse $9x^2 + 16y^2 = 144$ which pass through the point $(2, 3)$ are

Let $S$ and $S'$ be the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and $P$ be a variable point on the ellipse. The maximum area of the triangle $PSS'$ is ............. square units.