Question
The locus of the orthocenter of the triangle formed by the
lines
The locus of the orthocenter of the triangle formed by the
lines
$$\eqalign{
& \left( {1 + p} \right)x - py + p\left( {1 + p} \right) = 0, \cr
& \left( {1 + q} \right)x - qy + q\left( {1 + q} \right) = 0, \cr} $$
and $$y=0,$$ where $$p \ne q,$$ is :
A.
a hyperbola
B.
a parabola
C.
an ellipse
D.
a straight line
Answer :
a straight line
Solution :
The triangle is formed by the lines
$$\eqalign{ & AB:\left( {1 + p} \right)x - py + p\left( {1 + p} \right) = 0 \cr & AC:\left( {1 + q} \right)x - qy + q\left( {1 + q} \right) = 0 \cr & BC:y = 0 \cr} $$
So that the vertices are
$$A\left( {pq,\left( {p + 1} \right)\left( {q + 1} \right)} \right),\,\,B\left( { - p,\,0} \right),\,\,C\left( { - q,\,0} \right)$$
Let $$H\left( {h,\,k} \right)$$ be the orthocenter of $$\Delta ABC.$$ Then as $$AH \bot BC$$ and passes through $$A\left( {pq,\left( {p + 1} \right)\left( {q + 1} \right)\,} \right)$$
The equation of $$AH$$ is $$x=pq$$
$$\therefore h = pq.....(1)$$
Also $$BH$$ is perpendicular to $$AC$$
$$\eqalign{ & \therefore {m_1}{m_2} = - 1 \Rightarrow \frac{{k - 0}}{{h + p}} \times \frac{{1 + q}}{q} = - 1 \cr & \Rightarrow \frac{k}{{pq + p}} \times \frac{{1 + q}}{q} = - 1\,\,\,\,\,\,\left( {{\text{using equation (1)}}} \right) \cr & \Rightarrow k = - pq.....(2) \cr} $$
From (1) and (2) we observe $$h+k=0$$
$$\therefore $$ Locus of $$\left( {h,\,k} \right)$$ is $$x+y=0$$ which is a straight line.
The triangle is formed by the lines
$$\eqalign{ & AB:\left( {1 + p} \right)x - py + p\left( {1 + p} \right) = 0 \cr & AC:\left( {1 + q} \right)x - qy + q\left( {1 + q} \right) = 0 \cr & BC:y = 0 \cr} $$
So that the vertices are
$$A\left( {pq,\left( {p + 1} \right)\left( {q + 1} \right)} \right),\,\,B\left( { - p,\,0} \right),\,\,C\left( { - q,\,0} \right)$$
Let $$H\left( {h,\,k} \right)$$ be the orthocenter of $$\Delta ABC.$$ Then as $$AH \bot BC$$ and passes through $$A\left( {pq,\left( {p + 1} \right)\left( {q + 1} \right)\,} \right)$$
The equation of $$AH$$ is $$x=pq$$
$$\therefore h = pq.....(1)$$
Also $$BH$$ is perpendicular to $$AC$$
$$\eqalign{ & \therefore {m_1}{m_2} = - 1 \Rightarrow \frac{{k - 0}}{{h + p}} \times \frac{{1 + q}}{q} = - 1 \cr & \Rightarrow \frac{k}{{pq + p}} \times \frac{{1 + q}}{q} = - 1\,\,\,\,\,\,\left( {{\text{using equation (1)}}} \right) \cr & \Rightarrow k = - pq.....(2) \cr} $$
From (1) and (2) we observe $$h+k=0$$
$$\therefore $$ Locus of $$\left( {h,\,k} \right)$$ is $$x+y=0$$ which is a straight line.