Question

lf the pair lines $a{x}^2+2(a+b)xy+b{y}^2=0$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sectors, then

• A. $3a^2+2ab+3b^2=0$
• B. $3a^2-2ab+3b^2=0$
• C. $3a^2-10ab+3b^2=0$
• D. $3a^2+10ab+3b^2=0$

Answer 1

The correct option is A $3a^2+2ab+3b^2=0$

Let $AOB$ and $COD$ be the two given diameter lines which divide the circle into four sectors such that

Area $AOD=3$Area of $BOD$ (given)

$\Rightarrow \angle BOD=\frac{\pi }{4}$

$\Rightarrow$ Angle between the pair of given lines $=\frac{\pi }{4}$

$\tan \frac{\pi }{4}=\frac{2\sqrt{{(a+b)}^2-ab}}{a+b}$

$\Rightarrow 1=\frac{2\sqrt{a^2+b^2+ab}}{a+b}\tan \frac{\pi }{4}=\frac{2\sqrt{{(a+b)}^2-ab}}{a+b}\Rightarrow 1=\frac{2\sqrt{a^2+b^2+ab}}{a+b}$

$\Rightarrow 3a^2+3b^2+2ab=0$