Suppose the pairs of straight lines $2 x^2+a x y+3 y^2=0$ and $2 x^2+b x y-3 y^2=0$ are such that they have…
Suppose the pairs of straight lines $2 x^2+a x y+3 y^2=0$ and $2 x^2+b x y-3 y^2=0$ are such that they have one common line with the other two remaining perpendicular. Then the values of $a$ and $b$ respectively are
$-5,1$
$5,-1$
5,1
$5, \frac{1}{5}$
Solution
For $a x^2+2 h x y+b y^2=0$ ...(i)
If the slope of 2 lines representing ...(i)
are $m_1$ and $m_2$ then
$\begin{aligned} & m_1+m_2=-2 h / b \\ & m_1 \cdot m_2=a / b\end{aligned}$
For $2 x^2+a x y+3 y^2=0$ ...(ii)
If slope of common line is $m$ and slope of other 2 are $m_1$ and $m_2$ respectively
$\begin{aligned} & m+m_1=-a / 3, m . m_1=2 / 3 \\ & m+m_2=\mathrm{b} / 3, m \cdot m_2=-2 / 3 \\ & m m_1=-m m_2 \\ & m_1=-m_2\end{aligned}$
other 2 lines are perpendicular
$\begin{aligned} & m_1 \cdot m_2=-1 \\ & m_2=-1 / m_1 \\ & m_1=-m_2=-\left(\frac{-1}{m_1}\right) \\ & m_1^2=1 \\ & m_1= \pm 1 \\ & \therefore m_2=\mp 1\end{aligned}$
For $m_1=1$ and $m_2=-1$
$\begin{aligned} & m=2 / 3 \\ & m+m_1=-\frac{a}{3}=\frac{2}{3}+1=\frac{5}{3} \\ & a=-5 \\ & m+m_2=\frac{b}{3}=\frac{2}{3}-1=\frac{-1}{3} \\ & b=-1\end{aligned}$
$(a, b) \equiv(-5,-1)$
for $m_1=-1, m_2=1$
$\begin{aligned} & m=-2 / 3 \\ & m+m_1=\frac{-2}{3}-1=-\frac{5}{3}=-\frac{a}{3} \Rightarrow a=5 \\ & m+m_2=-\frac{2}{3}+1=\frac{1}{3}=\frac{b}{3} \Rightarrow b=1 \\ & (a, b) \equiv(5,1)\end{aligned}$