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The locus of the orthocentre of the triangle formed by the lines
$ (1+p) x-p y+p(1+p)=0, $
$ (1+q) x-q y+q(1+q)=0,$
and $y=0$, where $p \neq q$, is
a hyperbola
a parabola
an ellipse
a straight line
Solution
The correct option is $D y=-x$
Let $L_1:(1+p) x-p y+p(1+p)=0$,
$L_2:(1+q) x-q y+q(1+q)=0$ and
$L_3: y=0$
The coordinates of vertices are
$A(-p, 0), B(-q, 0), C(p q,(1+p)(1+q))$
The equation of the altitude through $C$ is $x=p q \cdots(1)$
The equation of the altitude through $B$ is $px +(1+ p ) y +\lambda=0$
Putting $B(-q, 0)$, we get
$-p q+\lambda=0 \Rightarrow \lambda=p q$
Now, equation of altitude through $B$
$p x+(1+p) y+p q=0 \cdots$
Solving equation $(1)$ and $(2)$, we get
$p^2 q+(1+p) y+p q=0$
$\Rightarrow y=-p q$
Let $(h, k)$ be the coordinates of the orthocentre
$h=p q, k=-p q$
$\Rightarrow k=-h$
Hence, the locus of orthocentre is $y=-x$.
