A circle with centre at the origin and radius equal to a meets the axis of x at A and B. P( α) and Q( β ) are two points on this circle so that α • β = 2γ , where γ is a constant. The locus of the point of intersection of AP and BQ is -
Text Solution
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Coordinates of A are (–a, 0) and of P are (a cos α , a sin α )
∴ Equation of AP is y = 
(x + a)
or y = tan ( α /2) (x + a) … (i)
Similarly equation of BQ is y = 
(x – a)
or y = – cot ( β /2) (x – a) … (ii)
we now eliminate α , β from (i) and (ii)
From (i) and (ii) tan ( α /2) = 
, tan ( β /2) = 
Now α – β = 2 γ
⇒ tan γ = 
= 
⇒ tan γ = 
= 
⇒ x 2 + y 2 – 2ay tan γ = a 2
which is the required locus
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