A circle whose centre coincides with the origin having radius a cuts the X-axis at A and B . If P and Q are two points on the circle whose parametric angles differ by 2 θ , then the locus of the intersection point of AP and BQ, is -
Text Solution
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Let P ≡ (a cos α , a sin α ) and Q ≡ (a cos β , a sin β )
where β – α = 2 θ
Also, A ≡ (a, 0) and B ≡ (–a, 0)
If R(h, k) be the intersection point of AP and BQ, then
slope of AR = slope of AP [ R lies on AP]
i.e. 
= 
i.e. tan 
= 
… (1)
and slope of BR = slope of BQ [ R lies on BQ]
i.e. 
= 
i.e. tan 
= 
… (2)
Since, β – α = 2 θ , we have
– 
= θ
i.e. 
= tan θ [from equations (1) and (2)]
i.e. 
= tan θ
i.e. h 2 + k 2 – 2ak tan θ = a 2
Hence, the locus of R is
x 2 + y 2 – 2ay tan θ = a 2 .
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