If alpha x^2+2 gamma x y+beta y^2=0 is the eq | vedclass

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(D) The given pair of lines is $b h x^2 + a b x y + a h y^2 = 0$. To find the pair of lines perpendicular to this,we replace $x$ with $y$ and $y$ with

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$\frac{4 h^2}{a b}$

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(D) The given pair of lines is $b h x^2 + a b x y + a h y^2 = 0$. To find the pair of lines perpendicular to this,we replace $x$ with $y$ and $y$ with $-x$ in the equation. Substituting $x 	o y$ and $y 	o -x$,we get: $b h (y)^2 + a b (y)(-x) + a h (-x)^2 = 0$ $b h y^2 - a b x y + a h x^2 = 0$ Rearranging the terms,we get: $a h x^2 - a b x y + b h y^2 = 0$. Comparing this with $\alpha x^2 + 2 \gamma x y + \beta y^2 = 0$,we have: $\alpha = a h$,$\beta = b h$,and $2 \gamma = -a b \implies \gamma = -\frac{a b}{2}$. Now,calculate $\frac{\alpha \beta}{\gamma^2}$: $\frac{\alpha \beta}{\gamma^2} = \frac{(a h)(b h)}{(-\frac{a b}{2})^2} = \frac{a b h^2}{\frac{a^2 b^2}{4}} = \frac{4 a b h^2}{a^2 b^2} = \frac{4 h^2}{a b}$. Thus,the correct option is $D$.

If $\alpha x^2+2 \gamma x y+\beta y^2=0$ is the equation of a pair of lines passing through the origin and perpendicular to the pair of lines $b h x^2+a b x y+a h y^2=0$ $(a \neq 0, b \neq 0)$,then $\frac{\alpha \beta}{\gamma^2}=$

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