JEE Advance - Mathematics (2009 - Paper 2 Offline - No. 19)
The locus of the 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 :
$$(1 + p)x - py + p(1 + p) = 0, $$
$$(1 + q)x - qy + q(1 + q) = 0$$
and $$y = 0$$, where $$p \ne q$$, is :
a hyperbola.
a parabola.
an ellipse.
a straight line.
Explanation
The intersection point of $$y = 0$$ with first line is $$B( - p,0)$$.
The intersection point of $$y=0$$ with second line is $$A( - q,0)$$.
The intersection point of the two lines is $$C(pq,(p + 1)(q + 1))$$.
The altitude from C to AB is $$x = pq$$.
The altitude from B to AC is
$$y = - {q \over {1 + q}}(x + p)$$
Solving these two equations, we get $$x = pq$$ and $$y = - pq$$.
Hence, the locus of orthocentre is $$x + y = 0$$.
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