Let the ellipse 3x^2 + py^2 = 4 pass through | vedclass

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(C) The equation of the circle is $x^2 + y^2 - 2x - 4y - 11 = 0$. The centre $C$ is $(1, 2)$ and the radius $r = \sqrt{1^2 + 2^2 - (-11)} = \sqrt{1 +

Vedclass · Accepted Answer

$70$

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(C) The equation of the circle is $x^2 + y^2 - 2x - 4y - 11 = 0$. The centre $C$ is $(1, 2)$ and the radius $r = \sqrt{1^2 + 2^2 - (-11)} = \sqrt{1 + 4 + 11} = 4$. Since the ellipse $3x^2 + py^2 = 4$ passes through $(1, 2)$,we have $3(1)^2 + p(2)^2 = 4$,which gives $3 + 4p = 4$,so $p = \frac{1}{4}$. The ellipse equation is $3x^2 + \frac{1}{4}y^2 = 4$,or $\frac{x^2}{4/3} + \frac{y^2}{16} = 1$. Here $a^2 = \frac{4}{3}$ and $b^2 = 16$. Since $b^2 > a^2$,this is a vertical ellipse. The eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{4/3}{16}} = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}}$. The focal distances of a point $(x_0, y_0)$ on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $b \pm ey_0$. Thus,$f_1 = b + ey_0 = 4 + \sqrt{\frac{11}{12}} 	imes 2$ and $f_2 = b - ey_0 = 4 - \sqrt{\frac{11}{12}} 	imes 2$. Then $f_1 f_2 = (4)^2 - (\sqrt{\frac{11}{12}} 	imes 2)^2 = 16 - (\frac{11}{12} 	imes 4) = 16 - \frac{11}{3} = \frac{48 - 11}{3} = \frac{37}{3}$. Finally,$6f_1f_2 - r = 6(\frac{37}{3}) - 4 = 2(37) - 4 = 74 - 4 = 70$.

Let the ellipse $3x^2 + py^2 = 4$ pass through the centre $C$ of the circle $x^2 + y^2 - 2x - 4y - 11 = 0$ with 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

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