Question

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

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

Answer 1

The correct option is D A straight line

Given, lines are $(1+p)x-py+p(1+p)=0$ ....(i)
and $(1+q)x-qy+q(1+q)=0$ ....(ii)
On solving Eqs. (i) and (ii), we get
$C\{pq,(1+p\left)\right(1+q\left)\right\}$
$\therefore$ Equation of altitude CM passing through C and perpendicular to AB is
$x=pq$ ....(iii)
$\because$ Slope of line (ii) is$\left(\frac{1+q}{q}\right)$
$\therefore$ slope of altitude BN (as shown in figure ) is $\frac{-q}{1+q}$


$\therefore$ Equation of BN is $y-0=\frac{-q}{1+q}(x+p)$
$\Rightarrow y=\frac{-q}{(1+1)}(x+p)$ ....(iv)
Let orthocentre triangle be H(h, k), which is the point of intersection of Eqs. (iii) and (iv).
On solving Eqs. (iii) and (iv), we get
$x=pqandy=-pq\Rightarrow h=pqandk=-pq\therefore h+k=0$
$\therefore$ Locus of H(h,k) is x+y=0 which is a straight line.