question_answer

If 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) \[3{{a}^{2}}-10ab+3{{b}^{2}}=0\]

B) \[3{{a}^{2}}-2ab+3{{b}^{2}}=0\]

C) \[3{{a}^{2}}+10ab+3{{b}^{2}}=0\]

D) \[3{{a}^{2}}+2ab+3{{b}^{2}}=0\]

Correct Answer: D

Solution :

[d]

As per question area of one sector = 3 area of another sector

\[\Rightarrow \] Angle at centre by one sector \[=3\times \]angle at centre by another sector

Let one angle be \[\theta \] then other \[=3\theta \]

Clearly \[\theta +3\theta =180\Rightarrow \theta =45{}^\circ \]

\[\therefore \] Angle between the diameters represented by combined equation

\[a{{x}^{2}}+2(a+b)xy+b{{y}^{2}}=0\] is \[45{}^\circ \]

\[\therefore \]Using \[\tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}\]

we get \[\tan 45{}^\circ =\frac{2\sqrt{{{(a+b)}^{2}}-ab}}{a+b}\]

\[\Rightarrow 1=\frac{2\sqrt{{{a}^{2}}+{{b}^{2}}+ab}}{a+b}\]

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

\[\Rightarrow {{a}^{2}}+{{b}^{2}}+2ab=4{{a}^{2}}+4{{b}^{2}}+4ab\]

\[\Rightarrow 3{{a}^{2}}+3{{b}^{2}}+2ab=0\]