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

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

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

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

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

If the foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} – \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide, then the value of ${b^2}$ is

Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis, respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ ad $E_2$ at $P, Q$ and $R$, respectively. Suppose that $P Q=P R=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression$(s)$ is(are)

$(A)$ $e_1^2+e_2^2=\frac{43}{40}$

$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$

$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$

$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points $(4,-1)$ and $(-2, 2)$ is

A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$