Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ from any of its foci?

The eccentricity of an ellipse is $2/3$, latus rectum is $5$ and centre is $(0, 0)$. The equation of the ellipse is

Find the equation for the ellipse that satisfies the given conditions: Centre at $(0,\,0),$ major axis on the $y-$ axis and passes through the points $(3,\,2)$ and $(1,\,6)$

Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right), y_1<0, y_2<0$, be the end points of the latus rectum of the ellipse $x^2+4 y^2=4$. The equations of parabolas with latus rectum $P Q$ are

$(A)$ $x^2+2 \sqrt{3} y=3+\sqrt{3}$

$(B)$ $x^2-2 \sqrt{3} y=3+\sqrt{3}$

$(C)$ $x^2+2 \sqrt{3} y=3-\sqrt{3}$

$(D)$ $x^2-2 \sqrt{3} y=3-\sqrt{3}$

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-

The equations of the common tangents to the ellipse, $ x^2 + 4y^2 = 8 $ $\&$  the parabola $y^2 = 4x$  can be