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Q.

If the pair of lines ax2+2(a+b)xy+by2=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

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a

3a2-10ab+3b2=0

b

3a2-2ab+3b2=0

c

3a2+10ab+3b2=0

d

3a2+2ab+3b2=0

answer is D.

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Detailed Solution

Let one angle be θ then other =3θ Clearly θ+3θ=180⇒θ=45° Angle between the diameters represented by combined equationax2+2(a+b)xy+by2=0 is 45°∴ Using tanθ=2h2-aba+b⇒1=2a2+b2+aba+b⇒(a+b)2=4a2+b2+ab ⇒3a2+2ab+3b2=0
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If the pair of lines ax2+2(a+b)xy+by2=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