If one of the lines given by the equation $2 x^2+a x y+3 y^2=0$ coincide with one of those given by the…

If one of the lines given by the equation $2 x^2+a x y+3 y^2=0$ coincide with one of those given by the equation $2 x^2+b x y-3 y^2=0$, while the other two lines are perpendicular to each other, then the values of $a$ and $b$ are
  1. $a=-5$ and $b=1$
  2. $a=-4$ and $b=-1$
  3. $a=4$ and $b=1$
  4. $a=-5$ and $b=-1$

Solution

Given, equation of pair of straight line $2 x^2+a x y+3 y^2=0$ and $2 x^2+b x y-3 y^2=0$ Let the slope of line $m$ and $m_1$ of the equation $2 x^2+a x y+3 y^2=0$ $\therefore \quad m+m_1=\frac{-a}{3}$ and $m m_1=\frac{2}{3}$ Slope of the other line is $m$ and $m_2$. Then, $m+m_2=\frac{b}{3}$ and $m m_2=-2 / 3$ Given, $\quad m_1 m_2=-1$ ...(i) $\therefore \quad \frac{m m_1}{m m_2}=-1$ $\Rightarrow \quad m_1=-m_2$ $\Rightarrow \quad m_1+m_2=0$ ...(ii) From Eqs. (i) and (ii), we get $m_1=1$ and $m_2=-1$ $\therefore \quad m=2 / 3$ $m+m_1=-a / 3$ $\Rightarrow \quad \frac{2}{3}+1=-a / 3 \Rightarrow a=-5$ and $m+m_2=b / 3$ $\Rightarrow \quad \frac{2}{3}-1=b / 3 \Rightarrow b=-1$ $\therefore \quad a=-5$ and $b=-1$

Asked in: AP EAMCET 2021 (24 Aug Shift 2)

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