Find the equation of the hyperbola with eccentricity $e = 3/2$ and foci at $(\pm 2, 0)$.

Let $P(x_0, y_0)$ be the point on the hyperbola $3x^2 - 4y^2 = 36$ which is nearest to the line $3x + 2y = 1$. Then $\sqrt{2}(y_0 - x_0)$ is equal to:

If the normals drawn to the hyperbola $xy=4$ at $(\alpha_i, \beta_i)$ for $i=1, 2, 3, 4$ are concurrent at the point $(a, b)$,then $\frac{(\alpha_1+\alpha_2+\alpha_3+\alpha_4)}{(\beta_1+\beta_2+\beta_3+\beta_4)}(\alpha_1 \alpha_2 \alpha_3 \alpha_4) =$

If a directrix of a hyperbola centered at the origin and passing through the point $(4, -2\sqrt{3})$ is $5x = 4\sqrt{5}$ and its eccentricity is $e$,then

At which point does the line $2x + \sqrt{6}y = 2$ touch the hyperbola $x^2 - 2y^2 = 4$?

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 $(x^{2}+y^{2})^{2} = \alpha x^{2}+\beta y^{2}$,then $\alpha+\beta$ is equal to