Coordinates

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Here the actual coordinates of B and C do not matter, but their distance (6) does. So let us shift the origin to their midpoint (o) as shown and exploit the symmetry.

$\underline{Case 1} Symmetric case : B = C$

$4 \tan^2 \frac{B}{2} = 1$ giving $\tan \frac{B}{2} = \frac{1}{2}$ and $\tan B = \frac{4}{3}$ Giving OA1 = 4 = Semi Minor axis

$\underline{Case 2} : C = 90$

This makes $\tan \frac{C}{2} = 1$ giving $\tan \frac{B}{2} = \frac{1}{4}$ and $\tan B = \frac{8}{15}$ Giving $OA2 = \frac{48}{15}$

Substitute point $A2 = (3. 3.2)$ and b = 4 in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ gives a = 5

hence a.b = 4 x 5 = 20

Finally, an observation. BOA1 is a 3-4-5 right triangle Giving BA1 = 5 = Semi major axis. Hence B and C must be the foci too!

Let Semi-Major axis be 'p' and Semi-Minor axis be 'q' . For the Triangle ABC, we are given the co-ordinates B (2,0) and C (8,0). It is also given that $\ 4tan \dfrac{B}{2} tan\dfrac{C}{2} = 1$ --eqn (1)

First Lets use equation (1) using the half angle property of Triangles. We know that $\tan\dfrac{B}{2} = \sqrt{\dfrac{(s-a)(s-c)}{s(s-b)}}$ --eqn(2)
$\tan\dfrac{C}{2} = \sqrt{\dfrac{(s-a)(s-b)}{s(s-c)}}$ --eqn(3)

Where 's' is the semi-perimeter of trangle ie. $\dfrac{a+b+c}{2}$ side BC = a = 6 (Given)

         AB = c

         AC = b

Now, Multiplying equations (2) and (3) we get,

$\dfrac{(s-a)}{s}$

Therefore, $\ 4tan \dfrac{B}{2} tan\dfrac{C}{2} =4 \dfrac{(s-a)}{s} = 1$ Now substituting the value of s we get $4 \dfrac{(b+c-a)}{(a+b+c)} = 1$

Implies $5a = 3b + 3c$

substituting value of 'a' = 6 we get, $10 = b + c$

Since BC is fixed, B and C are the focus of the ellipse. Therefore, $BC = 2(p)(e)$ Let center of BC be 'O' . If $b = c$ , AO is the semi minor axis. So then b=c=5.

BO = 3 and AB = 5 (AOB is right angled at O) Hence, AO = 4. Implies q = 4

Since $(p)(e) = 3$ , and Q = 4,
$e^2 = 1 - \dfrac{q^2}{p^2}$

$p^2 e^2 = 9$

$p^2 - q^2 = 9$

$p^2 = 9 + 16$

$p = 5$

Finally, Since we require $pq$

$pq = 20$