The graph of the conic $ x^2 – (y – 1)^2 = 1$  has one tangent line with positive slope that passes through the origin. the point of tangency being $(a, b). $ Then  Eccentricity of the conic is

Let the hyperbola $H : \frac{ x ^{2}}{ a ^{2}}- y ^{2}=1$ and the ellipse $E: 3 x^{2}+4 y^{2}=12$ be such that the length of latus rectum of $H$ is equal to the length of latus rectum of $E$. If $e_{ H }$ and $e_{ E }$ are the eccentricities of $H$ and $E$ respectively, then the value of $12\left( e _{ H }^{2}+ e _{ E }^{2}\right)$ is equal to

Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse, $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b<5)$ and the hyperbola $\frac{ x ^{2}}{16}-\frac{ y ^{2}}{ b ^{2}}=1$ respectively satisfying $e _{1} e _{2}=1 .$ If $\alpha$ and $\beta$ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $(\alpha, \beta)$ is equal to

At the point of intersection of the rectangular hyperbola $ xy = c^2 $ and the parabola $y^2 = 4ax$  tangents to the rectangular hyperbola and the parabola make an angle $ \theta $ and $ \phi $ respectively with the axis of $X$, then

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $9 y^{2}-4 x^{2}=36$