Question

A circle with centre at the origin and radius equal to a meets the axis of x at A and B. P(a) and Q(b) are two points on this circle so that a – b = 2g, where g is a constant. The locus of the point of intersection of AP and BQ is –

• A. x² – y² – 2ay tan g = a²
• B. x² + y² – 2ay tan g = a²
• C. x² + y² + 2ay tan g = a²
• D. x² – y² + 2ay tan g = a²

Answer 1

Answer: (B)

x² + y² – 2ay tan g = a²

Answer 2

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

(a cos a, a sin a)

\ Equation of AP is y = $\frac{a\sin\alpha}{a(\cos\alpha + 1)}$ (x + a)

or y = tan (a/2) (x + a) … (i)

Similarly equation of BQ is y = $\frac{a\sin\beta}{a(\cos\beta - 1)}$ (x – a)

or y = – cot (b/2) (x – a) … (ii)

we now eliminate a, b from (i) and (ii)

From (i) and (ii) tan (a/2) = $\frac{y}{a + x}$, tan (/2) = $\frac{a - x}{y}$

Now a – b = 2g

Ž tan g = $\frac{\tan(\alpha/2) - \tan(\beta/2)}{1 + \tan(\alpha/2)\tan(\beta/2)}$

= $\frac{\frac{y}{a + x} - \frac{a - x}{y}}{1 + \frac{y}{a + x}.\frac{a - x}{y}}$

Ž tan g = $\frac{y^{2} - (a^{2} - x^{2})}{(a + x)y + (a - x)y}$= $\frac{x^{2} + y^{2} - a^{2}}{2ay}$

Ž x² + y² – 2ay tan g = a²

which is the required locus