The locus of the orthocentre of the triangle formed by the lines (1 + p)x – qy + q(1+q) = 0and y=0 where p is not equal to q isa hyperbolaa parabola\tan ellipse\ta straight line

let the orthocentre be H(h,k)
after plotting the line, points on x and y axis are:

-q(q +1)/2(p+1),0                 and   0,q+1   respectively.

Hence the bisector eqn. for sides x = 0  and y = 0  are ----------->

x = -q(q +1)/2(p + 1)     and   y = (q +1)/2

there fore pints of intesection is h,k that is x,y
equating and dividing them we get,

h/k  =  -q/(p+1)
or
y = (-(p+1)/q)x

or  y = mx
which is a straight line.