Find the length of the common chord of the ellipse $\frac{(x - 2)^2}{9} + \frac{(y + 2)^2}{4} = 1$ and the circle $x^2 + y^2 - 4x + 2y + 4 = 0$.

If $e$ and $e'$ are the eccentricities of the ellipse $5x^2 + 9y^2 = 45$ and the hyperbola $5x^2 - 4y^2 = 45$ respectively,then $ee' = $

An ellipse intersects the hyperbola $2x^2 - 2y^2 = 1$ orthogonally. The eccentricity of the ellipse is the reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes,then:
$(A)$ Equation of ellipse is $x^2 + 2y^2 = 2$
$(B)$ The foci of ellipse are $(\pm 1, 0)$
$(C)$ Equation of ellipse is $x^2 + 2y^2 = 4$
$(D)$ The foci of ellipse are $(\pm \sqrt{2}, 0)$

The quadratic equation whose roots are $l$ and $m$,where $l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^2 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)}$,is:

The equation of the chord joining two points $(x_1, y_1)$ and $(x_2, y_2)$ on the rectangular hyperbola $xy = c^2$ is:

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ coincide with the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2$ is equal to