The equation of the conic with focus at $(1, -1)$,directrix along $x - y + 1 = 0$ and with eccentricity $e = \sqrt{2}$ is:

The locus of the midpoints of the parallel chords with gradient $m$ of the rectangular hyperbola $xy = c^2$ is

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:

Let the ellipse $E: \frac{x^{2}}{144}+\frac{y^{2}}{169}=1$ and the hyperbola $H: \frac{x^{2}}{16}-\frac{y^{2}}{\lambda^{2}}=-1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$,then the value of $24(e+L)$ is:

Let the latus rectum of the hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ subtend an angle of $\frac{\pi}{3}$ at the centre of the hyperbola. If $b^2$ is equal to $\frac{l}{m}(1+\sqrt{n})$,where $l$ and $m$ are co-prime numbers,then $l^2+m^2+n^2$ is equal to . . . . . . .

The eccentricity of the hyperbola $4x^2 - 9y^2 = 36$ is