Conic Sections Question 202
Question: 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
Options:
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 $
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Answer:
Correct Answer: D
Solution:
[d] As per question area of one sector = 3 times area of another sector
$ \Rightarrow $ Angle at centre by one sector $ =\frac{1}{3} \times $ angle at centre by another sector Let one angle be $ \theta $ then other $ =3\theta $ Clearly $ \theta +3\theta =180^\circ \Rightarrow \theta =45^\circ $
$ \therefore $ Angle between the diameters represented by the combined equation $ ax^{2}+2(a+b)xy+by^{2}=0 $ is $ 45^\circ $
$ \therefore $ Using $ \tan \theta =\frac{2\sqrt{h^{2}-ab}}{a+b} $ we get $ \tan 45{}^\circ =\frac{2\sqrt{{{(a+b)}^{2}}-4ab}}{a+b} $
$ \Rightarrow 1=\frac{2\sqrt{a^{2}+b^{2}-ab}}{a+b} $
$ \Rightarrow {{(a+b)}^{2}}=a^{2}+2ab+b^{2}$
$ \Rightarrow a^{2}+b^{2}+2ab=(a+b)^{2}$
$ \Rightarrow 3a^{2}+3b^{2}+2ab=0 $
BETA
