Question

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

• A. 1
• B. $-1$
• C. 3
• D. $-3$

Answer 1

Answer: (D)

The pair of lines is $6x^2-xy+4c{y}^2=0.$ 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 $6x^2-xy+4c{y}^2=0$ $\therefore$ $6x^2-x\left(\frac{-3x}{4}\right)+4c{\left(\frac{-3x}{4}\right)}^2=0$ $\Rightarrow$ $6x^2+\frac{3x^2}{4}+4c\frac{9x^2}{16}=0$ $\Rightarrow$ $x^2(24+3+9c)=0$ $\Rightarrow$ $9c=-27$ $\Rightarrow$ $c=-3$