If the slope of one of the lines in the pair of lines $8 x^2+a x y+y^2=0$ is thrice the slope of the second…
- $8 \sqrt{\frac{2}{3}}$
- $6$
- $16 \sqrt{2}$
- $3 \frac{\sqrt{2}}{5}$
Solution
Let $m_1$ and $m_2$ be two slopes Since given equation is homogeneous $\Rightarrow m_1+m_2=\frac{-a}{1}$ and $m_1 m_2=8 ;$ Since $m_1=3 m_2$ $\therefore 3 m_2+m_2=-\mathrm{a} \Rightarrow 4 m_2=-\mathrm{a}$ ...(i) and $3 m_2^2=8 \Rightarrow m_2= \pm \sqrt{\frac{8}{3}}$ So, from (i), $-4 \sqrt{\frac{8}{3}}=-a \Rightarrow a=8 \sqrt{2 / 3}$
Asked in: AP EAMCET 2024 (22 May Shift 2)