Let the ellipse $3 x^2+\mathrm{py}^2=4$ pass through the centre $C$ of the circle $x^2+y^2-2 x-4 y-11=0$ of radius $r$. Let $f_1, f_2$ be the focal distances of the point C on the ellipse. Then $6 f_1 f_2-r$ is equal to

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Aicharya · Accepted Answer

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

Let the ellipse $3 x^2+\mathrm{py}^2=4$ pass through the centre $C$ of the circle $x^2+y^2-2 x-4 y-11=0$ of…

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