Question

lf the pair of 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 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

Answer: (D)

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

For given condition, angle between the lines will be ${45}^o$ 0r ${135}^o$

$\Rightarrow \tan {45}^o\text{or}\tan {135}^o=\frac{2\sqrt{(a+b{)}^2-ab}}{a+b}=1$

Squaring both sides,

$\Rightarrow (a+b{)}^2=4(a^2+b^2+ab)$

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