The equation of the bisectors of the angles b | vedclass

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(D) The equation of the curve is $x^2+xy+y^2+x+3y+1=0$ and the line is $x+y+2=0$. To find the pair of lines joining the origin to the intersection poi

Vedclass · Accepted Answer

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

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(D) The equation of the curve is $x^2+xy+y^2+x+3y+1=0$ and the line is $x+y+2=0$. To find the pair of lines joining the origin to the intersection points,we homogenize the equation of the curve using the line equation $\frac{x+y}{-2}=1$. 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 denominator: $4x^2+4xy+4y^2 - 2(x^2+4xy+3y^2) + (x^2+2xy+y^2) = 0$ $4x^2+4xy+4y^2 - 2x^2-8xy-6y^2 + x^2+2xy+y^2 = 0$ $3x^2-2xy-y^2 = 0$ This is the pair of lines $ax^2+2hxy+by^2=0$ where $a=3, h=-1, b=-1$. The equation of the angle bisectors is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$. $\frac{x^2-y^2}{3-(-1)} = \frac{xy}{-1}$ $\frac{x^2-y^2}{4} = \frac{xy}{-1}$ $-x^2+y^2 = 4xy$ $x^2+4xy-y^2 = 0$ Thus,the correct option is $D$.

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 straight line $x+y+2=0$ is

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