The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9 x^2+16 y^2=144$ is

The foci of $16{x^2} + 25{y^2} = 400$ are

The normal at $\left( {2,\frac{3}{2}} \right)$ to the ellipse, $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$ touches a parabola, whose equation is

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies

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

On the ellipse $4{x^2} + 9{y^2} = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are