If theta is the acute angle between the lines | vedclass

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(B) The equation of the curve is $x^2+xy+y^2+x+3y+1=0$ and the line is $x+y+2=0$. To homogenize the equation of the curve using the line,we write the

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$\frac{1}{\sqrt{5}}$

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(B) The equation of the curve is $x^2+xy+y^2+x+3y+1=0$ and the line is $x+y+2=0$. To homogenize the equation of the curve using the line,we write the line as $\frac{x+y}{-2}=1$. Substituting this into the curve equation: $x^2+xy+y^2+(x+3y)(\frac{x+y}{-2}) + 1(\frac{x+y}{-2})^2=0$. Multiplying by $4$ to clear the denominators: $4x^2+4xy+4y^2-2(x^2+4xy+3y^2)+(x^2+2xy+y^2)=0$. $4x^2+4xy+4y^2-2x^2-8xy-6y^2+x^2+2xy+y^2=0$. $3x^2-2xy-y^2=0$. Comparing this with $ax^2+2hxy+by^2=0$,we have $a=3, h=-1, b=-1$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$. $	an 	heta = \left| \frac{2\sqrt{(-1)^2-3(-1)}}{3-1} ight| = \left| \frac{2\sqrt{4}}{2} ight| = 2$. Since $	an 	heta = 2$,we have $\cos 	heta = \frac{1}{\sqrt{1+	an^2 	heta}} = \frac{1}{\sqrt{1+4}} = \frac{1}{\sqrt{5}}$.

If $\theta$ is the acute angle between the lines joining the origin to the points of intersection of the curve $x^2+xy+y^2+x+3y+1=0$ and the straight line $x+y+2=0$,then $\cos \theta=$

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