The curve represented by $\frac{x^2}{12-\alpha} + \frac{y^2}{\alpha-10} = 1$ is

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$.

If $x+2y+k=0, k>0$ is a tangent to the ellipse $2x^2+y^2=2$,then the equation of the normal to the given ellipse at $\left(\frac{1}{\sqrt{2}}, \frac{k}{3}\right)$ is:

An ellipse having foci at $(3, 3)$ and $(-4, 4)$ and passing through the origin has eccentricity equal to

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $36 x^{2}+4 y^{2}=144$.

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: