Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(ab)$ be given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, $?\left(t\right)=\frac{5}{12}+t-t^2$ , then $a^2+b^2$ is equal to:

Let $P\left(\mathrm{3,3}\right)$ be a point on the hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ . If the normal to it at $P$ intersects the $x$ -axis at $\left(\mathrm{9,0}\right)$ and $e$ is its eccentricity, then the ordered pair $\left(a^2,e^2\right)$ is equal to:

Let $x=4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$ . If $P(1,\beta ),\beta >0$ is a point on this ellipse, then the equation of the normal to it at $P$ is: