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 cruve as C varies over the ellipse. Find the area inside the curve.
+397 Let k be the x co-ordinate of the point C, then its y co-ordinate (C above the x axis) will be sqrt(1 - k^2/4).
(The path of H will be symmetric about the x axis, so we need only bother with the part above the x axis. We can double the calculated area to get the total area.)
The altitude through C will be the line x = k.
The altitude through B will be perpendicular to the line AC.
The slope of AC is (sqrt(1- k^2 /4) - 0)/(k + 4), so the slope of BH will be -(k + 4)/(sqrt(1-k^2/4) - 0).
The equation of BH will be
\(\displaystyle \frac{y-0}{x-4}=-\frac{k+4}{\sqrt{1-k^{2}/4}}\)
This intersects the altitude through C at H, so the y co-ordinate of H will be \(\displaystyle \frac{(4-k)(4+k)}{\sqrt{1-k^{2}/4}}=\frac{2(16-k^{2})}{\sqrt{4-k^{2}}}.\)
k varies from x = -2 to x = +2, so the area above the x axis determined by the path of H will be
\(\displaystyle \int^{2}_{-2}\frac{32-2x^{2}}{\sqrt{4-x^{2}}}dx.\)
Substitute \(x = 2\sin\theta,\)
and the integral becomes
\(\displaystyle \int^{\pi/2}_{-\pi/2}28+4\cos2\theta\:d\theta=28\pi.\)
That means that the total area will be (hopefully, check for mistakes,) \(56\pi.\)
