Question

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 2q, then the locus of the intersection point of AP and BQ, is –

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

Answer 1

Answer: (C)

x² + y² – 2ay tan q = a²

Answer 2

Let P ŗ (a cos a, a sin a) and Q ŗ (a cos b, a sin b)

where b – a = 2q

Also, A ŗ (a, 0) and B ŗ (–a, 0)

If R(h, k) be the intersection point of AP and BQ, then

slope of AR = slope of AP [Q R lies on AP]

i.e. $\frac { \sin \alpha } { \cos \alpha - 1 }$

i.e. tan$\left( \frac { \alpha } { 2 } \right)$ = … (1)

and slope of BR = slope of BQ [Q R lies on BQ]

i.e. $\frac { \sin \beta } { \cos \beta + 1 }$

i.e. tan$\left( \frac { \beta } { 2 } \right)$ = … (2)

Since, b – a = 2q, we have

$\frac { \beta } { 2 }$– $\frac { \alpha } { 2 }$ = q

i.e. $\frac { \tan \left( \frac { \beta } { 2 } \right) - \tan \left( \frac { \alpha } { 2 } \right) } { 1 + \tan \left( \frac { \beta } { 2 } \right) \tan \left( \frac { \alpha } { 2 } \right) }$= tan q [from equations (1) and (2)]

i.e. = tan q

i.e. h² + k² – 2ak tan q = a²

Hence, the locus of R is

x² + y² – 2ay tan q = a² .