The equation $\frac{x^2}{10-a} + \frac{y^2}{4-a} = 1$ represents an ellipse when:

At which points on the ellipse $4x^2 + 9y^2 = 1$ is the tangent line parallel to the line $8x = 9y$?

Let $A = \{(x, y) : y = mx + 1\}$,$B = \{(x, y) : x^2 + 4y^2 = 1\}$,and $C = \{(\alpha, \beta) : (\alpha, \beta) \in A \text{ and } (\alpha, \beta) \in B \text{ and } \alpha > 0\}$. If set $C$ is a singleton set,then the sum of all possible values of $m$ is:

What is the locus of the point of intersection of perpendicular tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$?

If the tangents on the ellipse $4x^2 + y^2 = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other,then $a^2$ is equal to

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