The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=4ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $2e^2=$

What is the equation of the common tangent to the circle $(x - 3)^2 + y^2 = 9$ and the parabola $y^2 = 4x$ above the $X$-axis?

$A$ common tangent $T$ to the curves $C_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $C_{2}: \frac{x^{2}}{42}-\frac{y^{2}}{143}=1$ does not pass through the fourth quadrant. If $T$ touches $C_{1}$ at $(x_{1}, y_{1})$ and $C_{2}$ at $(x_{2}, y_{2})$,then $|2x_{1} + x_{2}|$ is equal to $......$

If the curves $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ and $x^2 - y^2 = c^2$ intersect at right angles,then:

Let the hyperbola $H : \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ pass through the point $(2\sqrt{2}, -2\sqrt{2})$. $A$ parabola is drawn whose focus is the same as the focus of $H$ with positive abscissa,and the directrix of the parabola passes through the other focus of $H$. If the length of the latus rectum of the parabola is $e$ times the length of the latus rectum of $H$,where $e$ is the eccentricity of $H$,then which of the following points lies on the parabola?

The slopes of the common tangents to the parabola $(x - 1)^2 = 4(y - 2)$ and the ellipse $\frac{(x - 1)^2}{1} + \frac{(y - 2)^2}{2} = 1$ are $m_1$ and $m_2$. Then,$m_1^2 + m_2^2$ is equal to: