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Find the locus of the orthocenter of the triangle formed by the lines $(1 + p)x – py + p(1 + p) = 0$, $(1 + q)x – qy + q(1 + q) = 0, y = 0$ where $p\ne q$

My attempt is as follows:-

enter image description here Let orthocenter be $(h,k)$

Equation of perpendicular from $B$ to $AC$ will be $x=pq$

As orthocenter will lie on it, so $h=pq\tag{1}$

Equation of perpendicular from $C$ to $AB$ will be $(1+p)y+px+pq=0$

As orthocenter will lie on it, so $(1+p)k+ph+pq=0\tag{2}$

Equation of perpendicular from $A$ to $BC$ will be $(1+q)y+qx+pq=0$

As orthocenter will lie on it, so $(1+q)k+qh+pq=0\tag{3}$

Subtracting equation $(3)$ from $(2)$

$$(p-q)k+(p-q)h=0$$

As $p\ne q$,so $k+h=0$

So locus of orthocenter is $h+k=0$

But is this correct? Actual answer is just "straight line", so that's why I am not sure whether this is the correct locus or not.

user3290550
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2 Answers2

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I propose here to settle this question into a more general framework.

The point is that the three equations you have given belong to the same parametric family with general equation

$$\text{line} \ L_p : \ \ \ \ (1+p)x+(-p)y+p(1+p)=0 \tag{1}$$

(for the third one, take $p=-1$), .

Take a look at the following picture making obvious the fact that lines $L_p$ are exactly the tangents to a certain parabola (see remark below for a proof).

enter image description here

(one sees in particular that straight line with equation $y=0$ is one of these tangents).

Now, we can use a classical (well, rather classical) property of a parabola :

The orthocenter of the triangle determined by three tangents to a parabola lies on the so-called "directrix line" of the parabola with equation $x+y=0$ drawn in red on the picture (see for example property 7 of paragraph 4 in http://users.math.uoc.gr/~pamfilos/eGallery/problems/ParaProblems.htm#tri or lemma of page 200 in https://epub.ub.uni-muenchen.de/4550/1/Fritsch_Rudolf_4550.pdf (reference : Kiepert's parabola) or http://jwilson.coe.uga.edu/EMT669/Student.Folders/Giddings.Jemma/Parabolas/Parabolas.html).

Remarks :

1) All the facts given here can be established, starting from the fact that this parabola has implicit equation $(x-y)^2-2(x+y)+1=0$.

2) There is a somewhat similar "dual" property for the rectangular hyperbola : Prove: A triangle inscribed in a rectangular hyperbola has its orthocenter on that hyperbola ("dual" because it deals with chords instead of tangents).

Jean Marie
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The ortho-center is at $(pq,-pq)$ which lies on $x+y=0$ for any value of $p$ and $q$. So your answer is correct.

Z Ahmed
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