Let the ellipse 3x^2 + py^2 = 4 pass through the centre C of the circle x^2 + y^2 - 2x - 4…

Question

Let the ellipse $3x^2 + py^2 = 4$ pass through the centre $C$ of the circle $x^2 + y^2 - 2x - 4y - 11 = 0$ of radius $r$. Let $f_1, f_2$ be the focal distances of the point $C$ on the ellipse. Then $6f_1f_2 - r$ is equal to

Accepted Answer

$$\begin{aligned}
&E: \frac{x^2}{4 / 3}+\frac{y^2}{4 / P}=1\
&\text { Centre of circle }(1,2) \text {, radius }\
&\mathrm{r}=\sqrt{1+4+11}
\end{aligned}$$

$$\mathrm{r}=4$$

$\because$ E pass from centre $(1,2)$

$$\therefore \frac{3}{4}+\mathrm{P}=1$$

$\mathrm{P}=\frac{1}{4} \quad \therefore$ vertical ellipse

$$\begin{aligned}
& \mathrm{e}=\sqrt{1-\frac{4 / 3}{16}}=\sqrt{1-\frac{1}{12}}=\sqrt{\frac{11}{12}} \
& \therefore \text { Focal distance of } \mathrm{C}(\mathrm{~h}, \mathrm{k}) \
& =\mathrm{b} \pm \mathrm{ek} \
& \mathrm{~F}_1=4+\sqrt{\frac{11}{12}} \times 2 \
& \mathrm{~F}_2=4-\sqrt{\frac{11}{12}} \times 2 \
& \therefore \mathrm{~F}_1 \mathrm{~F}_2=16-\frac{11}{3}=\frac{37}{3}
\end{aligned}$$

$\therefore 6 \mathrm{~F}_1 \mathrm{~F}_2-\mathrm{r}=74-4=70$

Solutions · 1