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

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Solution

The correct option is D $- 3$
$6 x^{2} - x y + 4 c y^{2} = 0$
One solution is $y = \frac{- 3}{4} x$
$∴$ Let $t = \frac{y}{x}$
$4 c t^{2} - t + 6 = 0$
Substitute $t = \frac{- 3}{4}$ ,
$4 c \times \frac{9}{16} + \frac{3}{4} + 6 = 0$
$\Rightarrow c \times \frac{9}{4} + \frac{27}{4} = 0$
$\Rightarrow c = \frac{- 27}{9}$
$\Rightarrow c = - 3.$

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lf one of the lines given by 6x2−xy+4cy2=0 ,3x+4y=0, then c equals

The correct option is D −36x2−xy+4cy2=0One solution is y=−34x∴ Let t=yx4ct2−t+6=0Substitute t=−34,4c×916+34+6=0⇒c×94+274=0⇒c=−279⇒c=−3.

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