question_answer

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

A) a hyperbola

B) a parabola

C) an ellipse

D) a straight line

Correct Answer: D

Solution :

We have,
\[\left( 1+p \right)x-py+p\left( 1+p \right)=0\]	??. (i)
\[\left( 1+q \right)x-qy+q\left( 1+q \right)=0\]	?? (ii)
\[y=0\]	..?. (iii)
From Eqs. (i) and (ii), we get
\[x=pq,y=\left( p+1 \right)\left( q+1 \right)\]
From Eqs. (i) and (iii), we get
\[x=-p,y=0\]
From Eqs. (ii) and (iii), we get
\[x=-q,y=0\]

Equation of altitude from C to AB is

\[x=pq\]

Altitude from B to AC is

\[y=\frac{-q}{1+q}\left( x+p \right)\]

Put \[x=pq\], we get \[y=-pq\]

\[\because \,\,\,\,h=pq,k=-pq\]

\[\therefore \,\,h+k=0\]

Locus of orthocenter \[x=y=0\] which represent the equation of straight line.