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A circle with center at the origin and r...

A circle with center at the origin and radius equal to a meets the axis of `x` at `Aa n dB. P(alpha)` and `Q(beta)` are two points on the circle so that `alpha-beta=2gamma` , where `gamma` is a constant. Find the locus of the point of intersection of `A P` and `B Qdot`

Text Solution

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Let

The coordinates of A are ( -a,0) and that of P are `(a cos alpha, a sin alpha)`.
Therefore, the equation of AP is
`y=(a sin alpha)/(a(cos alpha+1))(x+a)`
or `y=tan.(alpha)/(2)(x+a)` (1)
Similarly, the equationof BQ is
`y=( a sin beta)/(a(cos beta-1))(x-a)`
or `y=- cot . (beta)/(2)(x-a)` (2)
We now eliminate `alpha` and `beta` from (1) and (2). Therefore,
`tan.(alpha)/(2)=(y)/(a+x),tan.(beta)/(2)=(a-x)/(y)`
Now, `alpha-beta=2gammaimpliesgamma=(alpha)/(2)-(beta)/(2)`
`:. tan gamma=(tan.(alpha)/(2)-tan.(beta)/(2))/(1+tan.(alpha)/(2)tan.(beta)/(2))=((y)/(a+x)-(a-x)/(y))/(1+(y)/(a+x).(a-x)/(y))`
or `tan gamma=(y^(2)-(a^(2)-x^(2)))/((a+x)y+(a-x)y)`
`=(x^(2)+y^(2)-a^(2))/(2ay)`
or `x^(2)+y^(2)-2ay tan gamma=a^(2)`
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