Find the equation of the ellipse whose axes are the coordinate axes,which passes through the point $(-3, 1)$ and has an eccentricity of $\sqrt{2/5}$.

$A$ point on the ellipse $4x^2 + 9y^2 = 36$,where the normal is parallel to the line $4x - 2y - 5 = 0$,is

The equation $\frac{x^2}{10-a} + \frac{y^2}{4-a} = 1$ represents an ellipse when:

The tangents drawn from the point $P(3, 4)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ touch the ellipse at points $A$ and $B$. The equation of the locus of a point which is equidistant from point $P$ and the line $AB$ is:

Let the ellipse $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, A < B$ have the same eccentricity $e = \frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$,and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$,then the area of the quadrilateral $ABCD$ equals:

The coordinate of a point on the auxiliary circle of the ellipse $x^{2}+2y^{2}=4$ corresponding to the point on the ellipse whose eccentric angle is $60^{\circ}$ will be