Looking for an alternate (sleeker) approach

Question

I stumbled upon a problem asked in IIT JEE Advanced 2009:

The locus of the orthocentre of the triangle formed by the lines $(1+p)x-py+p(1+p) = 0$,$(1+q)x - qy + q(1+q)$ and $y = 0$, where $ p \neq q$ is

I am aware of the approach that I have to find the point of intersection of any two lines in terms of $p$ and $q$.

Then I have to write the equation of line perpendicular to the third line passing through the other point. Similarly I have to do it for another pair of lines. Then I have to calculate the intersection point of the altitudes and then eliminate the parameters to get the locus, which eventually turns out to be a straight line. (After 4 attempts in which I missed some or the other thing in the intermediate steps)

However, this approach took me 3 pages to solve. And to the best of my knowledge, questions of IIT have some key point, using which the question turns out to be much easier. I am not sure if this has any. If you have some other approach/ ideas, plz let me know

Looking forward to some suggestions.

Answer 1

The locus depends on two parameters ($p$ and $q$) and I'd expect then it to be a surface. But only curves are mentioned among the answers: that means that one of the parameters is redundant and you can fix it at a value of your choice.

You can take, for instance, $q=-1$ and find the orthocenter of the resulting triangle. In that case the triangle degenerates into the segment with endpoints $A=(1,0)$ and $B=(-p,0)$; in approaching the limit, the altitude perpendicular to $x$-axis tends to the line $x=-p$, while the one perpendicular to the $q$-line will pass through $A$, and is thus $(1+p)y=-p(x-1)$. Their intersection can then be easily found as $H=(-p,p)$.

Answer 2

As you know the fact I just asked you in the comments, I am going to use it here by manipulating the equation of the family of lines.

Let us say $l:\left(1+t\right)x-ty+t\left(1+t\right)=0$ is tangent to a curve $S$ for all values of $t$. So, if you solve the line with the point $(x_1,y_1)$, the equation must yield only one value of $t$. Conversely, the locus of all points $A:(x,y)$ such that $t$ has two coincident roots when the line $l$ is solved with $A$, is nothing but $S$

The line can be written as $$t^{2}+\left(1+x-y\right)t+\left(x\right)=0$$

For coincident roots, $D=0$ $$\implies (1+x-y)^2=4x$$ is the required locus.

If you had studied about envelopes, you could have thought it yourselves. An interactive demonstration is this

So, now you know that the family is actually a family of tangent lines to a parabola. Hence, the orthocentre of the triangles formed by it always lie on a same line. :)