Question

The locus of orthocentre of the triangle formed by the 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 allipse
• D. a straight line

Answer 1

The correct option is D a straight line

Intersection point of $y=0$ with first line is $B(-p,0)$
Intersection point of $y=0$ with second line is $A(-q,0)$
Intersection point of the two lines is $C(pq,(p+1\left)\right(q+1)$
Altitude from $C$ to $AB$ is $x=pq$
Altitude from $B$ to $AC$ is $y=\frac{-q}{1+q}$$(x+p)$
solving the two lines we get $x=pq$ and $y=-pq$
$\therefore$ Locus of orthocenter is $x+y=0$
That is a straight line.