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If the pair of lines ax^{2}+2(a+b)xy+by^{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

  • Option 1)

    3a^{2}-2ab+3b^{2}=0

  • Option 2)

    3a^{2}-10ab+3b^{2}=0

  • Option 3)

    3a^{2}+2ab+3b^{2}=0

  • Option 4)

    3a^{2}+10ab+3b^{2}=0

 

Answers (1)

best_answer

As we learned 

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

Here a =a, b=b, h=a+b

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

where  \theta = \frac{180-\theta }{3}\Rightarrow \theta = 45^{\circ}

\Rightarrow 2\sqrt{\left ( a+b \right )^{2}-ab}= a+b

\Rightarrow 4a^2+4b^2+4ab=a^2+b^2+2ab

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


Option 1)

3a^{2}-2ab+3b^{2}=0

Option 2)

3a^{2}-10ab+3b^{2}=0

Option 3)

3a^{2}+2ab+3b^{2}=0

Option 4)

3a^{2}+10ab+3b^{2}=0

Posted by

Aadil

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