Question

Two straight lines rotate about two fixed points. If they start from their position of coincidence such that one rotates at the rate double that of the other. Prove that the locus of their point of intersection is a circle.

Answer 1

As usual we take the fixed points A and B to be at a distance 2aapart along x-axis and their mid-point be chosen as origin so that A (a, 0) and B (- a, 0). Initially the coincided along this line AB. These lines now rotate so that BP makes an angle 8 and AP makes an angle 2$\theta$ and both these meet at P, whose locus we have to find. If tan$\theta$ = m then tan$2\theta =\frac{2m}{1-m^2}$
Eq. of BP is y = m(x + a) $\therefore$ m=$\frac{y}{x+a}$
Eq. of A P is y = $\frac{2m}{1-m^2}$ (x - a)
In order to find the locus of their point of intersection, eliminate m between (1) and (2)
$[1-\frac{y^2}{(x+a{)}^2}]y=2.\frac{y}{x+a}$ (x - a)
$(x-a{)}^2-y^2$ = 2 (x + a ) (x - a)
or $x^2+2ax+a^2-y^2=2x^2-.2a^2$
or $x^2+y^2-2ax-3a^2$ = 0