Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ be an ellipse,whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:

Suppose $A, B, C$ and $D$ are the four intersection points of the curves $\frac{x^2}{18}+\frac{y^2}{8}=1$ and $x^2-y^2=5$ in $I, II, III$ and $IV$ quadrants respectively. If $\theta_1, \theta_2, \theta_3$ and $\theta_4$ respectively are the angles between the curves at $A, B, C$ and $D$, then

If the tangent to the parabola $y^2 = x$ at a point $(\alpha, \beta)$,$(\beta > 0)$ is also a tangent to the ellipse $x^2 + 2y^2 = 1$,then $\alpha$ is equal to

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The scalar product of the vectors $\vec{OA}$ and $\vec{OB}$ is:

The tangent at $P$ to a parabola $y^2 = 4ax$ meets the directrix at $U$ and the latus rectum at $V$. Then $\triangle SUV$ (where $S$ is the focus) :

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: