Question

The locus of the orthocentre of the triangle formed by the lines $(1 + p )x -p y + p (1 + p) = 0 (1+ q)x-qy + < 7(1+ q) = 0$ 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)

a straight line

Answer 2

Given, lines are (1 + p )x -p y + p (1 + p) =0\hspace25mm ... (i)
and \hspace25mm (1 + q)x-qy+ q (1 + q) = 0 \hspace25mm ...(ii)
C$\\{pq, (1 + p) (1 + q)\\}$
$\therefore$Equation of altitude CM passing through C and perpendicular to A B is
$\because$ Slope of line (ii) is \hspace25mm ... (iii)
$\therefore$ Slope of altitude BN (as shown in frigure) is $\frac{-q}{1+q}$
$\therefore$ Equation of B N is y - 0 $=\frac{-q}{1+q}(x+p)$
\Rightarrow \hspace25mm y=\frac{-q}{(1+q)}(x+p) \hspace 25mm...(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
\hspace25mm x = pq
and \hspace25mm y = -pq
\Rightarrow \hspace25mm h = pq
and \hspace25mm k = - pq
\therefore \hspace25mm h + k = 0
$\therefore$ Locus of H (h , k) is x+ y = 0