If the pair of lines $a x^2+2(a+b) x y+b y^2=0$ lie along diameters of a circle and divide the circle into…

If the pair of lines $a x^2+2(a+b) x y+b y^2=0$ 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
  1. $3 a^2-10 a b+3 b^2=0$
  2. $3 a^2-2 a b+3 b^2=0$
  3. $3 a^2+10 a b+3 b^2=0$
  4. $3 a^2+2 a b+3 b^2=0$

Solution

$ \begin{aligned} & \left|\frac{2 \sqrt{(a+b)^2-a b}}{a+b}\right|=1 \\ & \Rightarrow(a+b)^2=4\left(a^2+b^2+a b\right) \\ & \Rightarrow 3 a^2+3 b^2+2 a b=0 . \end{aligned} $

Asked in: JEE Main 2005

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