The conic represented by $x = 2(\cos t + \sin t), y = 5(\cos t - \sin t)$ 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

$A$ staircase of length $l$ rests against a vertical wall and a floor of a room. Let $P$ be a point on the staircase,nearer to its end on the wall,that divides its length in the ratio $1 : 2$. If the staircase begins to slide on the floor,then the locus of $P$ is

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse is:

The eccentricity of the ellipse $\left( \frac{x - 3}{y} \right)^2 + \left( 1 - \frac{4}{y} \right)^2 = \frac{1}{9}$ is

Let $f(x)=x^2+9$,$g(x)=\frac{x}{x-9}$,$a=f(g(10))$,and $b=g(f(3))$. If $e$ and $l$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{a}+\frac{y^2}{b}=1$,then $8e^2+l^2$ is equal to.