Question

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

• A. A hyperbola
• B. A parabola
• C. An ellipse
• D. A straight line

Answer 1

Answer: (D)

A straight line

A straight line.

Intersection point of y = 0 with first line is the point $B(-p,0)$

Intersection point of y = 0 with second line is the point $A(-q,0)$

Intersection point of the two lines is the point $C(pq,(p+1\left)\right(q+1\left)\right)$

Altitude from C to AB is $x=pq$

Altitude from B to AC is $y=\frac{-p}{1+p}(x+q)$

On solving these two we get $x=pq,y=-pq$

locus of orthocenter is $x+y=0$