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 (

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 (

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 (B) a parabola (C) an ellipse (D) a straight line
A. a hyperbola
B. a parabola
C. an ellipse
D. a straight line

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - D
The intersection point y=0 with the first line si B(-p, 0).
The intersection point of y=0 with the second line is A(-q, 0).
The intersection point of the two line is
C(pq, (p+1)(q+1)
The altitude from C to AB is x = pq.
The altitude from B to AC is
`y = -(q)/(1+q)(x+p)` Solving these two, we get x=pq and y=-pq.
Therefore, the locus of the orthocenter is x+y=0.