If the eccentricity of an ellipse is $1/\sqrt{2}$,then its latus rectum is equal to its

The eccentricity of the conic $36x^2 + 144y^2 - 36x - 96y - 119 = 0$ is

The equation of the ellipse whose focus is $(6, 7)$,directrix is $x + y + 2 = 0$,and eccentricity $e = 1/\sqrt{3}$ is:

For an ellipse with eccentricity $e = \frac{1}{2}$,the centre is at the origin. If one of its directrices is $x = 4$,then the equation of the ellipse is

For $0 < \theta < \frac{\pi}{2}$,four tangents are drawn at the four points $(\pm 3 \cos \theta, \pm 2 \sin \theta)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. If $A(\theta)$ denotes the area of the quadrilateral formed by these four tangents,the minimum value of $A(\theta)$ is

The curve represented by $x = 3(\cos t + \sin t)$ and $y = 4(\cos t - \sin t)$ is