If frac{x^2}{a} + frac{2xy}{h} + frac{y^2}{b} | vedclass

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(B) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Multiplying by $abh$,we get $bhx^2 + 2abyx + ahy^2 = 0$. Comparing this

Vedclass · Accepted Answer

$9 : 8$

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(B) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Multiplying by $abh$,we get $bhx^2 + 2abyx + ahy^2 = 0$. Comparing this with the standard form $Ax^2 + 2Hxy + By^2 = 0$,we have $A = bh$,$2H = 2ab \Rightarrow H = ab$,and $B = ah$. Let the slopes be $m_1$ and $m_2$. We are given $m_2 = 2m_1$. For the equation $Ax^2 + 2Hxy + By^2 = 0$,the sum of slopes $m_1 + m_2 = -\frac{2H}{B}$ and the product of slopes $m_1 m_2 = \frac{A}{B}$. $m_1 + 2m_1 = 3m_1 = -\frac{2ab}{ah} = -\frac{2b}{h} \Rightarrow m_1 = -\frac{2b}{3h}$. $m_1(2m_1) = 2m_1^2 = \frac{bh}{ah} = \frac{b}{a} \Rightarrow 2\left(-\frac{2b}{3h}ight)^2 = \frac{b}{a}$. $2 	imes \frac{4b^2}{9h^2} = \frac{b}{a} \Rightarrow \frac{8b^2}{9h^2} = \frac{b}{a}$. Dividing both sides by $b$ (assuming $b 
eq 0$),we get $\frac{8b}{9h^2} = \frac{1}{a} \Rightarrow \frac{ab}{h^2} = \frac{9}{8}$.

If $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$ represents a pair of straight lines and the slope of one of the lines is twice that of the other,then $ab : h^2$ is:

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