Let $x^2=4 k y, k>0$ be a parabola with vertex $A$. Let $B C$ be its latusrectum. An ellipse with centre on $B C$ touches the parabola at $A$, and cuts $B C$ at points $D$ and $E$ such that $B D=D E=E C(B, D, E, C$ in that order). The eccentricity of the ellipse is

From the point$ C(0,\lambda )$ two tangents are drawn to ellipse $x^2\ +\ 2y^2\ = 4$ cutting major axis at $A$ and $B$. If  area of $\Delta$ $ABC$ is minimum, then value of $\lambda$  is-

If the point of intersections of the ellipse $\frac{ x ^{2}}{16}+\frac{ y ^{2}}{ b ^{2}}=1$ and the circle $x ^{2}+ y ^{2}=4 b , b > 4$ lie on the curve $y^{2}=3 x^{2},$ then $b$ is equal to:

If $ \tan\  \theta _1. tan \theta _2 $ $= -\frac{{{a^2}}}{{{b^2}}}$  then the chord joining two points $\theta _1 \& \theta _2$  on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}$ $= 1$  will subtend a right angle at :

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

The pole of the straight line $x + 4y = 4$ with respect to ellipse ${x^2} + 4{y^2} = 4$ is