Let the circle $x^{2} + y^{2} - 8 x = 0$ and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points $A$ and $B$ , where $A$ is in first quadrant. If an ellipse is constructed with vertices at $A, B$ and having eccentricity $\frac{1}{\sqrt{3}}$ , then the distance between its foci is

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Solution

Given, $x^{2} + y^{2} - 8 x = 0 \dots (1)$
$\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1 \dots (2)$
Solving $(1)$ and $(2)$ ,
$x^{2} + 4 (\frac{x^{2}}{9} - 1) - 8 x = 0 \Rightarrow 9 x^{2} + 4 (x^{2} - 9) - 72 x = 0 \Rightarrow 13 x^{2} - 72 x - 36 = 0 \Rightarrow (x - 6) (13 x + 6) = 0 \Rightarrow x = 6 (∵ x \neq - \frac{6}{13})$
$∴ A \equiv (6, 2 \sqrt{3})$ and $B \equiv (6, - 2 \sqrt{3})$

Length of major axis of ellipse, $2 a = A B = 4 \sqrt{3}$
$e = \frac{1}{\sqrt{3}}$
$∴$ Distance between the foci $= 2 a e = 4$

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Q. Let

F_{1} (x_{1}, 0)

and

F_{2} (x_{2}, 0),

where

x_{1} < 0, x_{2} > 0,

be the foci of the ellipse

\frac{x^{2}}{9} + \frac{y^{2}}{8} = 1.

A parabola having vertex at the origin and focus at

F_{1}

intersects the ellipse at the point

M

in the second quadrant and at point

N

in the third quadrant. If the distance between

M

and

N

d

, then the value of

d^{2}

Let $F_{1} (x_{1}, 0)$ and $F_{2} (x_{2}, 0),$ where $x_{1} < 0$ and $x_{2} > 0$ be the foci of the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{8} = 1$ suppose a parabola having vertex at the origin and focus at $F_{2}$ intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.
The orthocentre of $Δ F_{1} M N$ is