Let the foci of the ellipse $$\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$$ and the hyperbola $$\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}$$ coincide. Then the length of the latus rectum of the hyperbola is :

Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. Let e' and l' respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $${e^2} = {{11} \over {14}}l$$ and $${\left( {e'} \right)^2} = {{11} \over 8}l'$$, then the value of $$77a + 44b$$ is equal to :

Let the eccentricity of the hyperbola $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ be $$\sqrt {{5 \over 2}} $$ and length of its latus rectum be $$6\sqrt 2 $$. If $$y = 2x + c$$ is a tangent to the hyperbola H, then the value of c² is equal to :

The normal to the hyperbola

$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$$ at the point $$\left( {8,3\sqrt 3 } \right)$$ on it passes through the point :