# Vertices of a variable triangle are (3, 4), (5cos theta, 5sin theta), (5sin theta, 5cos theta) where theta in R. Given that the orthocenter of this triangle traces a conic, evaluate its eccentricity.

Find the Locus of the Orthocenter
Vertices of a variable triangle are $\left(3,4\right)\phantom{\rule{0ex}{0ex}}\left(5\mathrm{cos}\theta ,5\mathrm{sin}\theta \right)\phantom{\rule{0ex}{0ex}}\left(5\mathrm{sin}\theta ,-5\mathrm{cos}\theta \right)$ where $\theta \in \mathbb{R}$. Given that the orthocenter of this triangle traces a conic, evaluate its eccentricity.
I was able to find the locus after three long pages of cumbersome calculation. I found the equations of two altitudes of this variable triangle using point slope form of equation of a straight and then solved the two lines to get the orthocenter. However, the equation turned out to be of a non standard conic. I evaluated its Δ to find that it's an ellipse, but I don't know how to find the eccentricity of a general ellipse.
Moreover, there must be a more elegant way of doing this since the questions in my worksheet are to be solved within 5 to 6 minutes each but this took way long using my approach.
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Step 1
Note that the vertices of the triangle lie on ${x}^{2}+{y}^{2}=25$ which is a circle of radius 5 units and centered at origin.
Step 2
Now, the circumcenter of this variable triangle is the origin, i.e, $\text{O}\equiv \left(0,0\right)$.
Also, the centroid of this variable triangle is $\text{G}\equiv \left(\frac{5\mathrm{sin}\theta +5\mathrm{cos}\theta +3}{3},\frac{5\mathrm{sin}\theta -5\mathrm{cos}\theta +4}{3}\right)$.
Finally, since $\text{OG}:\text{GH}=1:2$ where H is the orthocenter, we have $3\left(\frac{5\mathrm{sin}\theta +5\mathrm{cos}\theta +3}{3}-2×0\right)=x$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=5\mathrm{sin}\theta +5\mathrm{cos}\theta +3$
Step 3
Similarly, $y=5\mathrm{sin}\theta -5\mathrm{cos}\theta +4$ where x and y are the co-ordinates of H
Solving the above equations for $\mathrm{sin}\theta$ and $\mathrm{cos}\theta$, we have,
$\mathrm{sin}\theta =\frac{x+y-7}{10}$
$\mathrm{cos}\theta =\frac{x-y+1}{10}$
Thus, the locus is $\left(x+y-7{\right)}^{2}+\left(x-y+1{\right)}^{2}=100$ which, on expanding gives ${x}^{2}+{y}^{2}-6x-8y-25=0$ which is an equation of a circle.
Thus eccentricity $\overline{)e=0}$.
###### Did you like this example?
Alonzo Odom
Step 1
Note that all points lie on a circle ${x}^{2}+{y}^{2}+{5}^{2}=25$.
So the center must be (0,0). Also, the centroid is $\left(\frac{3+5\mathrm{sin}\theta +5\mathrm{cos}\theta }{3},\frac{4-5\mathrm{sin}\theta +5\mathrm{cos}\theta }{3}\right)$.
Step 2
As we know, O, G, H lie on Euler's line, and:
$\frac{OG}{GH}=\frac{1}{2}$