If the slope of one of the lines in the pair | vedclass

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(A) The given equation is $8x^2 + axy + y^2 = 0$. Let the slopes of the two lines be $m_1$ and $m_2$. For a homogeneous equation $Ax^2 + Bxy + Cy^2 =

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$8 \sqrt{\frac{2}{3}}$

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(A) The given equation is $8x^2 + axy + y^2 = 0$. Let the slopes of the two lines be $m_1$ and $m_2$. For a homogeneous equation $Ax^2 + Bxy + Cy^2 = 0$,the sum of slopes is $m_1 + m_2 = -B/C$ and the product of slopes is $m_1 m_2 = A/C$. Here,$A = 8$,$B = a$,and $C = 1$. So,$m_1 + m_2 = -a$ and $m_1 m_2 = 8$. Given that one slope is thrice the other,let $m_1 = 3m_2$. Substituting this into the product equation: $(3m_2) 	imes m_2 = 8$ $\Rightarrow 3m_2^2 = 8$ $\Rightarrow m_2^2 = 8/3$. Substituting into the sum equation: $3m_2 + m_2 = -a$ $\Rightarrow 4m_2 = -a$ $\Rightarrow m_2 = -a/4$. Squaring both sides: $m_2^2 = a^2/16$. Equating the two values of $m_2^2$: $a^2/16 = 8/3$ $\Rightarrow a^2 = 128/3$ $\Rightarrow a = \pm \sqrt{128/3} = \pm 8 \sqrt{2/3}$. Since the options provide the positive value,$a = 8 \sqrt{\frac{2}{3}}$.

If the slope of one of the lines in the pair of lines $8x^2 + axy + y^2 = 0$ is thrice the slope of the other line,then $a =$

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