Let C be a point that varies ont the ellipse
x^2/4 + y^2 = 1.
Let H be the orthocenter of triangle ABC, where A = (-4,0) and B = (4,0). Then H traces a closed curve as C varies over the ellipse. Find the area inside the curve.
Note that the equation of the ellipse can be rewritten as $\frac{x^2}{2^2}+\frac{y^2}{1^2}=1$. Thus, the semi-major axis of the ellipse is $2$ and the semi-minor axis is $1$.
First, we find the equation of the line passing through $C$ and the midpoint of $\overline{AB}$. The midpoint of $\overline{AB}$ is $M=(0,0)$, and the slope of $CM$ is $\frac{y-0}{x-0}=\frac{y}{x}$. Using point-slope form, the equation of the line is $y=\frac{y}{x}x$, or $x^2-y^2=0$.
Next, we find the equation of the altitude from $A$ to $\overline{BC}$. Since $AB$ is horizontal, the altitude is vertical. Therefore, the equation is simply $x=-4$.
The intersection of these two lines is the foot of the altitude from $A$ to $\overline{BC}$, which we call $D$. Solving the system of equations $x^2-y^2=0$ and $x=-4$, we get $D=(-4,-4)$.
Similarly, we find that the foot of the altitude from $B$ to $\overline{AC}$ is $E=(4,4)$.
Finally, the orthocenter $H$ is the intersection of altitudes $\overline{AD}$ and $\overline{BE}$. The equation of line $\overline{AD}$ is $x=-4$, and the equation of line $\overline{BE}$ is $x=4$. Thus, $H=(0,-2)$.
As $C$ varies over the ellipse, the orthocenter $H$ traces out a curve, and we are looking for the area inside this curve. Note that the curve is symmetric about the $x$-axis, since the original triangle $ABC$ is symmetric about the $x$-axis. Therefore, we can just find the area of the region above the $x$-axis and multiply by $2$.
The curve is bounded by the line $y=-2$ and the ellipse. To find the $x$-coordinates of the endpoints of the curve, we set $y=0$ in the equation of the ellipse: $\frac{x^2}{2^2}+0=1\implies x=\pm 2\sqrt{3}$. Therefore, the area we seek is \begin{align*}
2\left(\int_{-2\sqrt{3}}^{-4} 2\sqrt{1-\frac{x^2}{4}}\,dx + \int_{-4}^{4} 2\sqrt{1-\frac{x^2}{4}}-2\,dx+\int_{4}^{2\sqrt{3}} 2\sqrt{1-\frac{x^2}{4}}\,dx\right)\\
=2\left(2\int_0^{4}\sqrt{1-\frac{x^2}{4}}\,dx+\int_{-4}^{4}2\sqrt{1-\frac{x^2}{4}}-2\,dx\right)\\
=2\left(2\cdot2\int_0^{\frac{\pi}{2}}2\cos^2\theta\,d\theta+2\cdot2\int_0^{\frac{\pi}{2}}\sin^2\theta\,d\theta\right)\\
=2\left(4\int_0^{\frac{\pi}{2}}(1+\cos 2\theta)\,d\theta\right)\\
=8\left[\theta+\frac{1}{2}\sin(2\theta)\right]_0^{\frac{\pi}{2}}\\
=8\cdot\frac{\pi}{2}\\
=\boxed{4\pi}.
\end{align*}
To find the area inside the curve traced by H as C varies over the given ellipse, we need to find the equation of the curve traced by the orthocenter H and then integrate the area enclosed by this curve.
First, let's find the coordinates of point C on the ellipse. The equation of the ellipse is:
x^2/4 + y^2 = 1.
Since C is a point that varies on the ellipse, let's express its coordinates in terms of a parameter t:
C = (2cos(t), sin(t))
Next, let's find the equation of the line passing through A(-4,0) and C(2cos(t), sin(t)). The slope of this line, m_AC, is given by:
m_AC = (sin(t) - 0) / (2cos(t) + 4) = sin(t) / (2cos(t) + 4).
The equation of this line can be written as:
y - 0 = (sin(t) / (2cos(t) + 4))(x - (-4)),
y = (sin(t) / (2cos(t) + 4))(x + 4).
Similarly, let's find the equation of the line passing through B(4,0) and C(2cos(t), sin(t)). The slope of this line, m_BC, is given by:
m_BC = (sin(t) - 0) / (2cos(t) - 4) = sin(t) / (2cos(t) - 4).
The equation of this line can be written as:
y - 0 = (sin(t) / (2cos(t) - 4))(x - 4),
y = (sin(t) / (2cos(t) - 4))(x - 4).
Now, let's find the equations of the altitudes of triangle ABC. The altitude from A is perpendicular to line BC and passes through A. The slope of line BC, m_BC, is given by sin(t) / (2cos(t) - 4), so the slope of the altitude from A, m_AH, is the negative reciprocal of m_BC:
m_AH = -1 / (sin(t) / (2cos(t) - 4)) = -(2cos(t) - 4) / sin(t) = (4 - 2cos(t)) / sin(t).
The equation of this altitude can be written as:
y - 0 = ((4 - 2cos(t)) / sin(t))(x - (-4)),
y = ((4 - 2cos(t)) / sin(t))(x + 4).
Similarly, the equation of the altitude from B can be found by finding the negative reciprocal of the slope of line AC:
m_BH = -1 / (sin(t) / (2cos(t) + 4)) = -(2cos(t) + 4) / sin(t) = (4 + 2cos(t)) / -sin(t),
y = ((4 + 2cos(t)) / -sin(t))(x - 4).
Hence, the equations of the altitudes from A and B are:
y_AH = ((4 - 2cos(t)) / sin(t))(x + 4),
y_BH = ((4 + 2cos(t)) / -sin(t))(x - 4).
Now, let's find the coordinates of the orthocenter H(x_H, y_H), which is the intersection point of the altitudes. Equating the equations of the altitudes, we get:
((4 - 2cos(t)) / sin(t))(x + 4) = ((4 + 2cos(t)) / -sin(t))(x - 4).
Simplifying this equation, we get:
(4 - 2cos(t))(x + 4)(-sin(t)) = (4 + 2cos(t))(x - 4)(sin(t)).
Expanding both sides, we have:
-4sin(t)x - 8sin(t)cos(t) - 16sin(t) = 4sin(t)x - 8sin(t)cos(t) + 16sin(t).
The x terms cancel out:
-16sin(t) = 16sin(t).
Dividing both sides by 16sin(t), we get:
-1 = 1.
This equation is not satisfied for any value of t. Hence, point H does not exist for any position of C on the ellipse.
Therefore, the orthocenter H does not exist for any point C on the given ellipse. Thus, the area inside the curve traced by H is zero.