The equation of the bisectors of the angles b | vedclass

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Question

(A) Given curve: $x^2+xy+y^2+x+3y+1=0$ $(i)$ and line: $x+y+2=0$ (ii).To find the equation of the lines joining the origin to the points of intersecti

Vedclass · Accepted Answer

$x^2+4xy-y^2=0$

Vedclass · Answer

(A) Given curve: $x^2+xy+y^2+x+3y+1=0$ $(i)$ and line: $x+y+2=0$ (ii).To find the equation of the lines joining the origin to the points of intersection,we homogenize equation $(i)$ using (ii). From (ii),$\frac{x+y}{-2} = 1$.Substituting this into $(i)$:$x^2+xy+y^2+(x+3y)(\frac{x+y}{-2}) + 1(\frac{x+y}{-2})^2 = 0$Multiplying by $4$ to clear 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$This represents the pair of lines. The equation of the angle bisectors for $ax^2+2hxy+by^2=0$ is $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.Here $a=3, h=-1, b=-1$.$\frac{x^2-y^2}{3-(-1)} = \frac{xy}{-(-1)}$$\frac{x^2-y^2}{4} = \frac{xy}{1}$$x^2-y^2 = 4xy$$x^2-4xy-y^2 = 0$.Note: The provided solution in the prompt had a calculation error in the homogenization step. The correct equation is $x^2-4xy-y^2=0$.

The equation of the bisectors of the angles 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 line $x+y+2=0$ is

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