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 $\frac{1}{2} \cdot 3 \cdot \frac{191}{9 \sqrt{55}} = \boxed{\frac{191 \sqrt{55}}{54}}.$
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 $PQR$.
3 answers
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 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
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
\[\frac{1}{2} \cdot \left(\frac{128}{119}\sqrt{55} - (-\frac{128}{119}\sqrt{55})\right) \cdot 6 = \frac{128}{119} \cdot 6 = \boxed{\frac{768}{119}}.\]
Using the formula for the area of a triangle, the area of $\triangle PQR$ is
\[\frac{1}{2} \cdot \left(\frac{128}{119}\sqrt{55} - (-\frac{128}{119}\sqrt{55})\right) \cdot 6 = \frac{128}{119} \cdot 6 = \boxed{\frac{768}{119}}.\]