(B) The angle $\theta$ between the lines represented by the homogeneous equation $ax^2+2hxy+by^2=0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2

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(B) The angle $\theta$ between the lines represented by the homogeneous equation $ax^2+2hxy+by^2=0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.Given $\theta = \frac{\pi}{4}$,we have $\tan \frac{\pi}{4} = 1$.Thus,$1 = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.Squaring both sides,we get $1 = \frac{4(h^2-ab)}{(a+b)^2}$.$(a+b)^2 = 4h^2 - 4ab$.$a^2 + 2ab + b^2 = 4h^2 - 4ab$.$4h^2 = a^2 + 6ab + b^2$.

Class 11 Mathematics Pair of straight lines Angle between the pair of straight lines, Condition for parallel and perpendicular lines

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

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A
$(a+b)^2$
B
$a^2+6ab+b^2$
C
$(a-2b)(2a+b)$
D
$a^2-6ab+b^2$

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Pair of straight lines

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Angle between the pair of straight lines, Condition for parallel and perpendicular lines

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