The eccentricity of the hyperbola $5{x^2} – 4{y^2} + 20x + 8y = 4$ is

Let a line $L_{1}$ be tangent to the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ and let $L_{2}$ be the line passing through the origin and perpendicular to $L _{1}$. If the locus of the point of intersection of $L_{1}$ and $L_{2}$ is $\left(x^{2}+y^{2}\right)^{2}=$ $\alpha x^{2}+\beta y^{2}$, then $\alpha+\beta$ is equal to

Let $P \left( x _0, y _0\right)$ be the point on the hyperbola $3 x ^2-4 y ^2$ $=36$, which is nearest to the line $3 x+2 y=1$. Then $\sqrt{2}\left( y _0- x _0\right)$ is equal to :

If $\mathrm{e}_{1}$ and $\mathrm{e}_{2}$ are the eccentricities of the ellipse, $\frac{\mathrm{x}^{2}}{18}+\frac{\mathrm{y}^{2}}{4}=1$ and the hyperbola, $\frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1$ respectively and $\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)$ is a point on the ellipse, $15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k},$ then $\mathrm{k}$ is equal to 

Let $m_1$ and $m_2$ be the slopes of the tangents drawn from the point $P (4,1)$ to the hyperbola $H: \frac{y^2}{25}-\frac{x^2}{16}=1$. If $Q$ is the point from which the tangents drawn to $H$ have slopes $\left| m _1\right|$ and $\left| m _2\right|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$ axis, then $\frac{(P Q)^2}{\alpha \beta}$ is equal to $…………$.