A circle with centre at the origin and radius equal to a meets the axis of x and

Question

A circle with centre at the origin and radius equal to a meets the axis of x and A and B.

P (α)

and

Q (β)

are two points on this circle so that

α - β = 2 γ

, where

γ

is a constant. The locus of the point of intersection of AP and BQ is

Options:

A . x2−y2−2aytanγ=a2

B . x2+y2−2aytanγ=a2

C . x2+y2+2aytanγ=a2

D . x2−y2+2aytanγ=a2

Answer: Option B
:
B

A Circle With Centre At The Origin And Radius Equal To A Mee...

Coordinates of A are (–a, 0) and of P are

(a c o s α, a s i n α)

∴

Equation of AP is

y = \frac{a s i n α}{a (c o s α + 1)} (x + a) \Rightarrow y = t a n (\frac{α}{2}) (x + a) \to (1)

Similarly equation of BQ =

y = \frac{a s i n β}{a (c o s β - 1)} (x - a) \Rightarrow y = - c o t (\frac{β}{2}) (x - a) \to (2)

From (1) and (2),

t a n (\frac{α}{2}) = \frac{y}{α + x}, t a n (\frac{β}{2}) = \frac{a - x}{y}

Now

α - β = 2 γ

\Rightarrow t a n γ = \frac{t a n (α / 2) - t a n (β / 2)}{(a + x) y + (a - x) y} = \frac{\frac{y}{α + x} - \frac{a - x}{y}}{1 + \frac{y}{α + x} . \frac{a - x}{y}} \Rightarrow t a n γ = \frac{t a n (α / 2) - t a n (β / 2)}{1 + t a n (α / 2) t a n (β / 2)} = \frac{x^{2} + y^{2} - a^{2}}{2 a y} \Rightarrow x^{2} + y^{2} - 2 a y t a n γ = a^{2}

which is required locus.

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