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:-

The locus of the mid points of the chords of the hyperbola $\mathrm{x}^{2}-\mathrm{y}^{2}=4$, which touch the parabola $\mathrm{y}^{2}=8 \mathrm{x}$, is :

The locus of the point of intersection of the lines $(\sqrt{3}) kx + ky -4 \sqrt{3}=0$ and $\sqrt{3} x-y-4(\sqrt{3}) k=0$ is a conic, whose eccentricity is ………….

If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$ on the hyperbola, is equal to

The chord $ PQ $ of the rectangular hyperbola $xy = a^2$  meets the axis of $x$ at $A ; C $ is the mid point of $  PQ\ \& 'O' $ is the origin. Then the $ \Delta ACO$  is :