Let the ellipse E:x^24/3+y^24/P=1 pass through the centre of the circle C: x^2+y^2-2x-4y-1…

Question

Let the ellipse $E:\dfrac{x^2}{4/3}+\dfrac{y^2}{4/P}=1$ pass through the centre of the circle $C: x^2+y^2-2x-4y-11=0$ of radius $r$. Let $f_1,f_2$ be the focal distances of the point (the centre of $C$) on the ellipse. Then $6f_1f_2-r$ is equal to:

Accepted Answer

Circle: centre $(1,2)$, $r=\sqrt{1+4+11}=4$.

$E$ passes through $(1,2)$: $\dfrac{1}{4/3}+\dfrac{4}{4/P}=1 \Rightarrow \dfrac34+P=1\Rightarrow P=\dfrac14$, so $E:\dfrac{x^2}{4/3}+\dfrac{y^2}{16}=1$ --- a vertical ellipse with semi-major $b=4$, semi-minor$^2=4/3$.

$$e=\sqrt{1-\frac{4/3}{16}}=\sqrt{\frac{11}{12}}.$$

Focal distances of $(1,2)$: $f_{1,2}=b\pm ek=4\pm 2\sqrt{11/12}$.

$$f_1f_2=16-4e^2=16-\frac{11}{3}=\frac{37}{3},\qquad 6f_1f_2-r=74-4=70.$$

Solutions · 1