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Question

(C) The equation of the pair of straight lines joining the origin to the points of intersection is obtained by homogenizing the equation of the curve

Vedclass · Accepted Answer

$\pi /3$

Vedclass · Answer

(C) The equation of the pair of straight lines joining the origin to the points of intersection is obtained by homogenizing the equation of the curve using the line equation: $x^2 + y^2 = 4 \left( \frac{x\sqrt{3} + y}{2} ight)^2$ $x^2 + y^2 = 4 \left( \frac{3x^2 + y^2 + 2\sqrt{3}xy}{4} ight)$ $x^2 + y^2 = 3x^2 + y^2 + 2\sqrt{3}xy$ $2x^2 + 2\sqrt{3}xy = 0$ Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 2$,$h = \sqrt{3}$,and $b = 0$. If $\alpha$ is the angle between the lines,then $	an \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$. $	an \alpha = \frac{2\sqrt{(\sqrt{3})^2 - (2)(0)}}{2 + 0} = \frac{2\sqrt{3}}{2} = \sqrt{3}$. Therefore,$\alpha = \frac{\pi}{3}$.

The angle between the lines joining the origin to the points of intersection of the line $x\sqrt{3} + y = 2$ and the curve $x^2 + y^2 = 4$ is

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