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

The normal drawn at the point $\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{7} = 1$ intersects its major axis at the point

$A$ particle is travelling in a clockwise direction on the ellipse $\frac{x^2}{100} + \frac{y^2}{25} = 1$. If the particle leaves the ellipse at the point $(-8, 3)$ and travels along the tangent to the ellipse at that point,then the point where the particle crosses the $Y$-axis is:

The ellipse $x^2 + 4y^2 = 4$ is inscribed in a rectangle aligned with the coordinate axes,which in turn is inscribed in another ellipse that passes through the point $(4,0)$. Then the equation of the outer ellipse 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:

If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{4 a^{2}}=1$ and the coordinate axes is $kab$,then $k$ is equal to ..... .