The equation of an ellipse is
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.
Point $Q$ is at $(0,b).$ Points $P$ and $R$ lie on the ellipse so that $\overline{PR}$ is parallel to the $x$-axis, and sides $\overline{QR}$ and $\overline{PQ}$ contain foci $F_1$ and $F_2,$ respectively. If $a=8$ and $b=3$, then find the area of triangle $PQR$.
The center of the ellipse is at the origin, so the foci are $(\pm c,0)$ for some $c,$ where $c = \sqrt{a^2 - b^2} = \sqrt{55}.$ Hence, $F_1 = (-\sqrt{55},0)$ and $F_2 = (\sqrt{55},0).$
Note that $P$ is at $(-x,y)$ and $R$ is at $(x,y)$ for some $x.$ Then by the distance formula,
\[PR = 2x,\]so $x = \frac{PR}{2}.$ Also,
\[PF_1 = \sqrt{(x + \sqrt{55})^2 + y^2}\]and
\[PF_2 = \sqrt{(x - \sqrt{55})^2 + y^2}.\]Thus, by the definition of an ellipse,
\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\]Hence, $x^2 = \frac{a^2 b^2 - b^2 y^2}{a^2}.$ Then
\begin{align*}
PF_1^2 - PF_2^2 &= (x + \sqrt{55})^2 + y^2 - (x - \sqrt{55})^2 - y^2 \\
&= 4x \sqrt{55} \\
&= 2 \sqrt{55} PR.
\end{align*}By the Law of Cosines on triangle $PQR,$
\begin{align*}
QR^2 &= PR^2 + PQ^2 - 2 \cdot PR \cdot PQ \cdot \cos \angle PQR \\
&= PR^2 + PF_2^2 - 2 \cdot PR \cdot PF_2 \cdot \frac{x - \sqrt{55}}{PF_2} \\
&= PR^2 + PF_2^2 - 2 PR (x - \sqrt{55}) \\
&= PR^2 + (x - \sqrt{55})^2 + y^2 - 2 PR (x - \sqrt{55}) \\
&= 2 a^2 - 2 PR \sqrt{55}.
\end{align*}Therefore,
\[QR^2 - (PF_1^2 - PF_2^2) = 2 a^2 - 2 PR \sqrt{55} - 2 \sqrt{55} PR = 2(a^2 - \sqrt{55} PR) = 2b^2.\]Hence, $PR = \frac{1}{\sqrt{55}} \cdot (a^2 - \frac{b^2}{2}).$
Finally, triangle $PQR$ has height 3, and
\begin{align*}
PQ &= PR - QR \\
&= \frac{1}{\sqrt{55}} \cdot (a^2 - \frac{b^2}{2}) - \sqrt{a^2 - b^2} \\
&= \frac{191}{9 \sqrt{55}}.
\end{align*}Therefore, the area of triangle $PQR$ is $\frac{1}{2} \cdot 3 \cdot \frac{191}{9 \sqrt{55}} = \boxed{\frac{191 \sqrt{55}}{54}}.$
we have c = √55.
The line QR has equation
x/√55 + y/3 = 1
This intersects the ellipse at (128/119 √55, -27/119)
So the area of triangle PQR is
(27/119 + 3)(256/119 √55)/2 ≈ 25.74
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The center of the ellipse is at the origin $(0,0)$.
Since $Q$ is at $(0,b)$, and $b=3$, point $Q$ is at $(0,3)$.
We know that the coordinates of the foci of the ellipse are $(\pm c, 0)$, where $c = \sqrt{a^2 - b^2}$. Substituting $a = 8$, $b = 3$ into this equation, we get $c = \sqrt{8^2 - 3^2} = \sqrt{55}$. So the foci are located at $F_1 = (-\sqrt{55}, 0)$ and $F_2 = (\sqrt{55}, 0)$.
Since $P$ and $R$ lie on the ellipse and $\overline{PR}$ is parallel to the $x$-axis, the coordinates of $P$ and $R$ can be written as $(\pm x, y)$.
We also know that $\overline{QR}$ and $\overline{PQ}$ contain the foci $F_1$ and $F_2$. Therefore, the sum of the distances from $P$ and $R$ to $F_1$ is equal to the distance from $Q$ to $F_2$.
Using the distance formula, the sum of the distances from $P$ and $R$ to $F_1$ can be calculated as follows:
$$PF_1 + RF_1 = \sqrt{(x - (-\sqrt{55}))^2 + (y - 0)^2} + \sqrt{(x - (-\sqrt{55}))^2 + (y - 0)^2}$$
$$= 2\sqrt{(x + \sqrt{55})^2 + y^2}$$
The distance from $Q$ to $F_2$ can be calculated as follows:
$$QF_2 = \sqrt{(0 - \sqrt{55})^2 + (3 - 0)^2} = \sqrt{55 + 9} = \sqrt{64} = 8$$
Therefore, we have the equation:
$$2\sqrt{(x + \sqrt{55})^2 + y^2} = 8$$
Dividing both sides of the equation by 2, we get:
$$\sqrt{(x + \sqrt{55})^2 + y^2} = 4$$
Squaring both sides of the equation, we get:
$$(x + \sqrt{55})^2 + y^2 = 16$$
Expanding and simplifying the equation, we get:
$$x^2 + 2x\sqrt{55} + 55 + y^2 = 16$$
Since $x^2/a^2 + y^2/b^2 = 1$, we know that $x^2/64 + y^2/9 = 1$.
Equating the two equations, we get:
$$\frac{x^2}{64} + \frac{2x\sqrt{55}}{64} + \frac{55}{64} + \frac{y^2}{9} = 16$$
Multiplying both sides of the equation by $64$, we get:
$$x^2 + 2x\sqrt{55} + 55 + \frac{64y^2}{9} = 1024$$
Rearranging the equation, we get:
$$x^2 + 2x\sqrt{55} + \left(\frac{64y^2}{9} - 969\right) = 0$$
Since $x^2 + 2x\sqrt{55} + \left(\frac{64y^2}{9} - 969\right) = 0$ represents a quadratic equation, the discriminant must be equal to $0$ to obtain real solutions for $x$.
The discriminant of the quadratic equation is given by:
$$\text{Discriminant} = (2\sqrt{55})^2 - 4(1)\left(\frac{64y^2}{9} - 969\right)$$
$$= 220 - \frac{256y^2}{9} + 3876$$
$$= \frac{95}{9} - \frac{256y^2}{9}$$
Setting the discriminant to $0$, we get:
$$\frac{95}{9} - \frac{256y^2}{9} = 0$$
Simplifying the equation, we get:
$$95 - 256y^2 = 0$$
Therefore, $256y^2 = 95$ and $y^2 = \frac{95}{256}$. Taking the square root of both sides of the equation, we get $y = \pm \frac{\sqrt{95}}{16}$.
We now have the coordinates of $P$ and $R$ as $(x, \frac{\sqrt{95}}{16})$ and $(x, -\frac{\sqrt{95}}{16})$, respectively.
The $x$-coordinate of $F_1$ is $-\sqrt{55}$, so the distance from $P$ or $R$ to $F_1$ can be calculated as follows:
$$PF_1 = \sqrt{(x - (-\sqrt{55}))^2 + \left(\frac{\sqrt{95}}{16} - 0\right)^2} = \sqrt{(x + \sqrt{55})^2 + \frac{95}{256}}$$
Therefore, the area of triangle $PQR$ can be calculated as follows:
$$\text{Area of triangle }PQR = \frac{1}{2} \cdot \left(\text{base}\right) \cdot \left(\text{height}\right)$$
$$= \frac{1}{2} \cdot \left(PQ + QR\right) \cdot PF_1$$
$$= \frac{1}{2} \cdot \left(3 + 2\left(\sqrt{(x + \sqrt{55})^2 + \frac{95}{256}}\right)\right) \cdot \sqrt{(x + \sqrt{55})^2 + \frac{95}{256}}$$
Substituting the values of $x$ from the equations obtained earlier, the area of triangle $PQR$ can be found.
I believe there is a small mistake in your calculation. The line $QR$ intersects the ellipse at $\left(\frac{128}{119}\sqrt{55}, -\frac{27}{119}\right),$ which means the $x$-coordinate of $Q$ is $\frac{128}{119}\sqrt{55}$ and the $y$-coordinate is $-3.$ Therefore, the height of $\triangle PQR$ is $3 - (-3) = 6,$ not $3 + 3 = 6.$
Using the formula for the area of a triangle, the area of $\triangle PQR$ is
\[\frac{1}{2} \cdot \left(\frac{128}{119}\sqrt{55} - (-\frac{128}{119}\sqrt{55})\right) \cdot 6 = \frac{128}{119} \cdot 6 = \boxed{\frac{768}{119}}.\]