Question

If one of the lines given by $ 6{{x}^{2}}-xy+4c{{y}^{2}}=0 $ is $ 3x+4y=0, $ then c equals:

• A. 1
• B. $ -1 $
• C. 3
• D. $ -3 $

Answer 1

Answer: (D)

The pair of lines is $ 6{{x}^{2}}-xy+4c{{y}^{2}}=0. $ On comparing with $ a{{x}^{2}}+2hxy+b{{y}^{2}}=0, $ we get $ a=6,2h=-1,b=4c $ $ \therefore $ $ {{m}_{1}}+{{m}_{2}}=-\frac{2h}{b}=\frac{1}{4c} $ and $ {{m}_{1}}+{{m}_{2}}=\frac{a}{b}=\frac{6}{4c} $ One line of given pair of lines is $ 3x+4y=0 $ Slope of line $ =-\frac{3}{4}={{m}_{1}} $ (say) $ \therefore $ $ -\frac{3}{4}+{{m}_{2}}=\frac{1}{4c} $ $ \Rightarrow $ $ {{m}_{2}}=\frac{1}{4c}+\frac{3}{4} $ $ \therefore $ $ \left( -\frac{3}{4} \right)\left( \frac{1}{4c}+\frac{3}{4} \right)=\frac{6}{4c} $ $ \Rightarrow $ $ -\frac{3}{4}\left( \frac{1+3c}{4c} \right)=\frac{6}{4c} $ $ \Rightarrow $ $ -\frac{3}{4}\left( \frac{1+3c}{4c} \right)=\frac{6}{4c} $ $ \Rightarrow $ $ 1+3c=\frac{-6\times 4}{3} $ $ \Rightarrow $ $ 1+3c=-8 $ $ \Rightarrow $ $ 3x=-9\Rightarrow c=-3 $ Alternate Solution $ \because $ $ 3x+4y=0 $ is one of the two lines. Hence $ y=-\frac{3x}{4} $ will satisfy the equation $ 6{{x}^{2}}-xy+4c{{y}^{2}}=0 $ $ \therefore $ $ 6{{x}^{2}}-x\left( \frac{-3x}{4} \right)+4c{{\left( \frac{-3x}{4} \right)}^{2}}=0 $ $ \Rightarrow $ $ 6{{x}^{2}}+\frac{3{{x}^{2}}}{4}+4c\frac{9{{x}^{2}}}{16}=0 $ $ \Rightarrow $ $ {{x}^{2}}(24+3+9c)=0 $ $ \Rightarrow $ $ 9c=-27 $ $ \Rightarrow $ $ c=-3 $