Mathematics - Ellipse Question with Solution | TestHub

$\mathrm{E}:\frac{{\mathrm{x}}^2}{4/3}+\frac{{\mathrm{y}}^2}{4/\mathrm{P}}=1$
Centre of circle ( 1,2 ), radius
$\begin{array}{rl}& \mathrm{r}=\sqrt{1+4+11}\\ & \mathrm{r}=4\end{array}$
$\because$ E pass from centre $(1,2)$
$\therefore \frac{3}{4}+\mathrm{P}=1$
$\mathrm{P}=\frac{1}{4}\therefore$ vertical ellipse
$\begin{array}{rl}& \mathrm{e}=\sqrt{1-\frac{4/3}{16}}=\sqrt{1-\frac{1}{12}}=\sqrt{\frac{11}{12}}\\ & \therefore \text{ Focal distance of }\mathrm{C}(\mathrm{h},\mathrm{k})\\ & =\mathrm{b}\pm \mathrm{ek}\\ & {\mathrm{F}}_1=4+\sqrt{\frac{11}{12}}\times 2\\ & {\mathrm{F}}_2=4-\sqrt{\frac{11}{12}}\times 2\\ & \therefore {\mathrm{F}}_1{\mathrm{F}}_2=16-\frac{11}{3}=\frac{37}{3}\\ & \therefore 6{\mathrm{F}}_1{\mathrm{F}}_2-\mathrm{r}=74-4=70\end{array}$