Trigonometric EquationHard
Question
A tangent is drawn at a point P on the parabola y2 = 4x to intersect x-axis at Q. From Q a line is drawn perpendicular to PQ to intersect y-axis at R and rectangle PQRS is completed, the equation of locus of point S is -
Options
A.x(4 - x)2 + 8y2 = 0
B.x(4 - x)2 - 8y2 = 0
C.x(4 + x)2 + 8y2 = 0
D.x2(4 - x) - 8y2 = 0
Solution
Let P be (t2, 2t)
∴ PQ : ty = x + t2
∴ Q : (-t2, 0)
∴ RQ : y = - tx - t3
⇒ R = (0, - t3)
Let S be (h, k)
∴
& 
Eliminating t, we get
x(4 - x)2 - 8y2 = 0
∴ PQ : ty = x + t2
∴ Q : (-t2, 0)
∴ RQ : y = - tx - t3
⇒ R = (0, - t3)
Let S be (h, k)
∴
& Eliminating t, we get
x(4 - x)2 - 8y2 = 0
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