If a tangent to the hyperbola $x^2 - \frac{y^2}{3} = 1$ is also a tangent to the parabola $y^2 = 8x$,then the equation of such a tangent with a positive slope is:

Consider a circle with its centre lying on the focus of the parabola $y^2 = 2px$ such that it touches the directrix of the parabola. Then,a point of intersection of the circle and the parabola is

Let $a, b$ and $\lambda$ be positive real numbers. Suppose $P$ is an end point of the latus rectum of the parabola $y^2 = 4 \lambda x$,and suppose the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through the point $P$. If the tangents to the parabola and the ellipse at the point $P$ are perpendicular to each other,then the eccentricity of the ellipse is

For some $\theta \in (0, \frac{\pi}{2})$,let the eccentricity and the length of the latus rectum of the hyperbola $x^{2} - y^{2} \sec^{2} \theta = 8$ be $e_{1}$ and $l_{1}$,respectively,and let the eccentricity and the length of the latus rectum of the ellipse $x^{2} \sec^{2} \theta + y^{2} = 6$ be $e_{2}$ and $l_{2}$,respectively. If $e_{1}^{2} = e_{2}^{2}(\sec^{2} \theta + 1)$,then $(\frac{l_{1}l_{2}}{e_{1}e_{2}}) \tan^{2} \theta$ is equal to . . . . . . .

The product of the slopes of the common tangents to the ellipse $\frac{x^2}{32} + \frac{y^2}{8} = 1$ and the parabola $y^2 = 8x$ is:

Find the equation of the ellipse whose focus is $(-1, 1)$,eccentricity is $1/2$,and directrix is $x - y + 3 = 0$.