If the lines represented by 2x^2 + 6xy + y^2 | vedclass

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Question

(A) The equation of the pair of bisectors of the angle between the lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$

Vedclass · Accepted Answer

$	heta$

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(A) The equation of the pair of bisectors of the angle between the lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.For the first pair of lines $2x^2 + 6xy + y^2 = 0$,the bisectors are $\frac{x^2 - y^2}{2 - 1} = \frac{xy}{3} \implies 3x^2 - 3y^2 = xy \implies 3x^2 - xy - 3y^2 = 0$.For the second pair of lines $4x^2 + 18xy + y^2 = 0$,the bisectors are $\frac{x^2 - y^2}{4 - 1} = \frac{xy}{9} \implies 9x^2 - 9y^2 = 3xy \implies 3x^2 - xy - 3y^2 = 0$.Since both pairs of lines have the same pair of angle bisectors,they are equally inclined to each other.If the angle between $L_1$ and $l_1$ is $	heta$,then by the property of equal inclination,the angle between $L_2$ and $l_2$ is also $	heta$.

If the lines represented by $2x^2 + 6xy + y^2 = 0$ are $L_1$ and $L_2$,and the lines represented by $4x^2 + 18xy + y^2 = 0$ are $l_1$ and $l_2$,and the acute angle between $L_1$ and $l_1$ is $\theta$,then the acute angle between $L_2$ and $l_2$ is :-

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