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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≠q, is

A
A hyperbola
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B
A parabola
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C
An ellipse
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D
A straight line
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Solution

The correct option is D A straight line
Given, lines are (1+p)xpy+p(1+p)=0 ....(i)
and (1+q)xqy+q(1+q)=0 ....(ii)
On solving Eqs. (i) and (ii), we get
C{pq,(1+p)(1+q)}
Equation of altitude CM passing through C and perpendicular to AB is
x=pq ....(iii)
Slope of line (ii) is(1+qq)
slope of altitude BN (as shown in figure ) is q1+q


Equation of BN is y0=q1+q(x+p)
y=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=pq and y=pq h=pq and k=pq h+k=0
Locus of H(h,k) is x+y=0 which is a straight line.

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