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Question

(C) The formula for the acute angle $	heta$ between the pair of lines represented by $ax^2+2hxy+by^2=0$ is given by $	an 	heta = \left| \frac{2\sqr

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$a^2+6ab+b^2$

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(C) The formula for the acute angle $	heta$ between the pair of lines represented by $ax^2+2hxy+by^2=0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$.Given $	heta = \frac{\pi}{4}$,we have $	an \frac{\pi}{4} = 1$.Therefore,$1 = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$.Squaring both sides,we get $1 = \frac{4(h^2-ab)}{(a+b)^2}$.This implies $(a+b)^2 = 4h^2 - 4ab$.Expanding the left side,$a^2 + 2ab + b^2 = 4h^2 - 4ab$.Rearranging the terms,$4h^2 = a^2 + 6ab + b^2$.

If the acute angle between the lines given by $ax^2+2hxy+by^2=0$ is $\frac{\pi}{4}$,then $4h^2=$

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