Trigonometric EquationMedium
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
Options
A.a hyperbola
B.a parabola
C.an ellipse
D.a straight line
Solution
Intersection point of y = 0 with first line is B(-p, 0)
Intersection point of y = 0 with second line is A(-q, 0)
Intersection point of the two lines is C(pq, (p + 1)(q + 1))
Altitude from C to AB is x = pq
Altitude from B to AC is y = -
(x + p)
Solving these two we get x = pq and y = - pq
∴ locus of orthocentre is x + y = 0.
Intersection point of y = 0 with second line is A(-q, 0)
Intersection point of the two lines is C(pq, (p + 1)(q + 1))
Altitude from C to AB is x = pq
Altitude from B to AC is y = -
(x + p)Solving these two we get x = pq and y = - pq
∴ locus of orthocentre is x + y = 0.
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