Question

A circle with centre at the origin and radius equal to a meets the axis of x and A and B. $P\left(\alpha \right)$ and $Q\left(\beta \right)$ are two points on this circle so that $\alpha -\beta =2\gamma$, where $\gamma$ is a constant. The locus of the point of intersection of AP and BQ is

• A. $x^2-y^2-2aytan\gamma =a^2$
• B. $x^2+y^2-2aytan\gamma =a^2$
• C. $x^2+y^2+2aytan\gamma =a^2$
• D. $x^2-y^2+2aytan\gamma =a^2$

Answer 1

The correct option is B

$x^2+y^2-2aytan\gamma =a^2$

Coordinates of A are (–a, 0) and of P are $(acos\alpha ,asin\alpha )$
$\therefore$ Equation of AP is $y=\frac{asin\alpha }{a(cos\alpha +1)}(x+a)\Rightarrow y=tan\left(\frac{\alpha }{2}\right)(x+a)\to \left(1\right)$
Similarly equation of BQ = $y=\frac{asin\beta }{a(cos\beta -1)}(x-a)\Rightarrow y=-cot\left(\frac{\beta }{2}\right)(x-a)\to \left(2\right)$
From (1) and (2), $tan\left(\frac{\alpha }{2}\right)=\frac{y}{\alpha +x},tan\left(\frac{\beta }{2}\right)=\frac{a-x}{y}$
Now $\alpha -\beta =2\gamma$
$\Rightarrow tan\gamma =\frac{tan\left(\alpha /2\right)-tan\left(\beta /2\right)}{(a+x)y+(a-x)y}=\frac{\frac{y}{\alpha +x}-\frac{a-x}{y}}{1+\frac{y}{\alpha +x}.\frac{a-x}{y}}\Rightarrow tan\gamma =\frac{tan\left(\alpha /2\right)-tan\left(\beta /2\right)}{1+tan\left(\alpha /2\right)tan\left(\beta /2\right)}=\frac{x^2+y^2-a^2}{2ay}\Rightarrow x^2+y^2-2aytan\gamma =a^2$ which is required locus.