Question

If the pair of line $a{x}^2+2(a+b)xy+b{y}^2=0$ 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 sector then-

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

Answer 1

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

From given, we have,

the area of one of the sectors is thrice the area of another sector

$a{x}^2+2(a+b)xy+b{y}^2=0$

Let $\theta$ be the angle between the lines

$\therefore \tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$

$\tan \theta =\left|\frac{2\sqrt{(a+b{)}^2-ab}}{a+b}\right|$

given,

$3A=B$

now $\theta ={45}^{\circ }$

$\therefore \tan 45=\left|\frac{2\sqrt{(a+b{)}^2-ab}}{a+b}\right|$

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