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

If the line $x \cos \alpha + y \sin \alpha = p$ is normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then

The eccentricity of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4, 0)$ and $(4, 0)$. Then,its eccentricity is

If $x+y+n=0, n>0$ is a normal to the ellipse $x^2+3y^2=3$ and $x+my+3=0, m < 0$ is a tangent to the ellipse $x^2+5y^2=5$,then the point of intersection of these two lines satisfies the equation

The sum of the focal distances of any point on the conic $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is