Find the equation of the ellipse whose latus rectum is $10$ and the length of the minor axis is equal to the distance between the foci.

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 for two distinct values of $p$ the lines $y=x+p$ touch the ellipse $E: \frac{x^2}{16} + \frac{y^2}{9} = 1$ at the points $A$ and $B$. Let the line $y = x$ intersect $E$ at the points $C$ and $D$. Then the area of the quadrilateral $ABCD$ is equal to

The area of the region bounded by the curve $9x^2 + 4y^2 - 36 = 0$ is (in $\pi$)

For an ellipse with eccentricity $e = \frac{1}{2}$,the centre is at the origin. If one of its directrices is $x = 4$,then the equation of the ellipse is

The area (in sq. units) of the triangle formed by the tangent and normal at a point $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the $X$-axis is