The angle between the straight lines x^2+4xy+ | vedclass

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(C) The given equation of the pair of straight lines is $x^2+4xy+y^2=0$. Comparing this with the general equation $ax^2+2hxy+by^2=0$,we get $a=1$,$2h=

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The given equation of the pair of straight lines is $x^2+4xy+y^2=0$. Comparing this with the general equation $ax^2+2hxy+by^2=0$,we get $a=1$,$2h=4 \Rightarrow h=2$,and $b=1$. The angle $	heta$ between the pair of straight lines is given by the formula: $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$. Substituting the values,we get: $	an 	heta = \left| \frac{2\sqrt{2^2-(1)(1)}}{1+1} ight| = \frac{2\sqrt{4-1}}{2} = \sqrt{3}$. Since $	an 	heta = \sqrt{3}$,we have $	heta = 60^{\circ}$.

The angle between the straight lines $x^2+4xy+y^2=0$ is....... (in $^{\circ}$)

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