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 \[\tan 45{}^\circ =\frac{2\sqrt{{{(a+b)}^{2}}-ab}}{a+b}\] |
\[\Rightarrow 1=\frac{2\sqrt{{{a}^{2}}+{{b}^{2}}+ab}}{a+b}\] |
\[\Rightarrow {{(a+b)}^{2}}=4({{a}^{2}}+{{b}^{2}}+ab)\] |
\[\Rightarrow {{a}^{2}}+{{b}^{2}}+2ab=4{{a}^{2}}+4{{b}^{2}}+4ab\] |
\[\Rightarrow 3{{a}^{2}}+3{{b}^{2}}+2ab=0\] |
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