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

Answer: (D)

Given, lines are $(1+p)x-py+p(1+p)=\mathrm{0....}$ (i)
and $(1+q)x-qy+q(1+q)=\mathrm{0....}$(ii)
On solving Eqs. (i) and (ii), we get
$C\{pq,(1+p)(1+q)\}$
$\therefore$ 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 $y-0=\frac{-q}{1+q}(x+p)$
$\Rightarrow y=\frac{-q}{(1+q)}(x+p).....$(iv)
Let orthocentre of triangle be $H(h,k)$, which is the point of intersection of Eqs. (iii) and (iv).
On solving Eqs. (iii) and (iv), we get
$x=pq$
$y=-pq$
$\Rightarrow h=pq$
and $k=-pq$
$\therefore h+k=0$
$\therefore$ Locus of $H(h,k)$ is $x+y=0$.