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.

Marcelo Mullins 2022-07-21 Answered
Find the Locus of the Orthocenter
Vertices of a variable triangle are ( 3 , 4 ) ( 5 cos θ , 5 sin θ ) ( 5 sin θ , 5 cos θ ) where θ 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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Answers (2)

renegadeo41u
Answered 2022-07-22 Author has 9 answers
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, O ( 0 , 0 ).
Also, the centroid of this variable triangle is G ( 5 sin θ + 5 cos θ + 3 3 , 5 sin θ 5 cos θ + 4 3 ) .
Finally, since OG : GH = 1 : 2 where H is the orthocenter, we have 3 ( 5 sin θ + 5 cos θ + 3 3 2 × 0 ) = x
x = 5 sin θ + 5 cos θ + 3
Step 3
Similarly, y = 5 sin θ 5 cos θ + 4 where x and y are the co-ordinates of H
Solving the above equations for sin θ and cos θ, we have,
sin θ = x + y 7 10
cos θ = x y + 1 10
Thus, the locus is ( x + y 7 ) 2 + ( x y + 1 ) 2 = 100 which, on expanding gives x 2 + y 2 6 x 8 y 25 = 0 which is an equation of a circle.
Thus eccentricity e = 0 .
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Alonzo Odom
Answered 2022-07-23 Author has 4 answers
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 ( 3 + 5 sin θ + 5 cos θ 3 , 4 5 sin θ + 5 cos θ 3 ) .
Step 2
As we know, O, G, H lie on Euler's line, and:
O G G H = 1 2
x 0 = α 3 , y 0 = β 3 H  is  ( 3 + 5 sin θ + 5 cos θ 3 , 4 5 sin θ + 5 cos θ )
Squaring and adding, ( α 3 ) 2 25 + ( β 4 ) 2 25 = 2 Circle,  e = 0
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