Question

lf one of the lines given by the equation $2x^2+axy+3y^2=0$ coincide with one of those given by $2x^2+bxy-3y^2=0$ and the other lines represented by them be perpendicular, then

• A. $a=-5,b=1$
• B. $a=5,b=-1$
• C. $a=5,b=1$
• D. None of these

Answer 1

Answer: (C)

$a=5,b=1$

Given eq

$2x^2+axy+3y^2=0$---(1)

$2x^2+bxy-3y^2=0$---(2)

Let $y=m_{1}x$ and $y=m_{2}x$ are pair of line of eq (1)

Then $y^2-(m_1+m_2)xy+m_1{m}_2{x}^2=0$

$m_1+m_2=\frac{-a}{3}$----(3)

and $m_1{m}_2=\frac{2}{3}$-----(4)

Similarly

$y=m_{1}x$ and $y=\frac{1}{m_2}x$ are pair of eq (2)

$y^2-(m_1-\frac{1}{m_2})xy+\frac{m_1}{m_2}{x}^2=0$

$m_1-\frac{1}{m_2}=\frac{b}{3}$----(5)

and $\frac{m_1}{m_2}=\frac{2}{3}$-----(6)

$\Rightarrow m_1=\frac{2}{3}{m}_2$

from eq (4) and (6)

$\frac{2}{3}{m}_2{m}_2=\frac{2}{3}$

$m_2^2=1$

$m_2=1$ or $-1$

SO $m_1=\frac{2}{3}$ or $-\frac{2}{3}$

from eq (3)

$m_1+m_2=-\frac{a}{3}$

$1+\frac{2}{3}=-\frac{a}{3}$ or $-1-\frac{2}{3}=-\frac{a}{3}$

$\frac{5}{3}=-\frac{a}{3}$ or $-\frac{5}{3}=-\frac{a}{3}$

$a=-5$ or $5$

$\therefore b=1$ or $-1$