Question

The coordinates of points a and b are (ak, 0) and (a/k, 0) if a point p moves such that pa=(kpb) find the locus of p

Answer 1

locus of point P is circle of equation x² + y² = a²

co-ordinates of points A and B are (ak, 0) and (a/k, 0). a point P(x, y) moves such that PA = kPB

squaring both sides,

PA² = k²PB²

using distance formula,

PA² = (x - ak)² + (y - 0)² = (x - ak)² + y²

PB² = (x - a/k)² + y²

so, (x - ak)² + y² = k²[(x - a/k)² + y² ]

⇒(x - ak)² - k²(x - a/k)² = k²y² - y²

⇒(1 - k²)x² -2akx + 2(a/k)x × k² + a²k² - k²(a²/k²) = y²(k² - 1)

⇒(1 - k²)x² + a²k² - a² + y²(1 - k²) = 0

⇒(1 - k²)x² + (1 - k²)y² = a²(1 - k²)

⇒x² + y² = a² , this is an standard form of circle whose centre lies at origin.

hence, locus of point P is circle of equation x² + y² = a²

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