Line Pair ax²+2hxy+by²=0 parallel to the General Equation

If the Equations $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a Pair of Lines, then $ax^2+2hxy+by^2=0$ represent a Pair of Lines through the Origin Parallel to the above Pair.

Let $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a line pair. Then the left hand side can be resolved into two linear factors and the equation may be written as \[(l_1x+m_1y+n_1)(l_2x+m_2y+n_2)=0\] where $l_1l_2=a$, $m_1m_2=b$, $n_1n_2=c$, $l_1m_2+m_1l_2=2h$, $l_1n_2+n_1l_2=2g$, $m_1n_2+m_2n_1=2f$.

The separate equations of the lines are \[l_1x+m_1y+n_1=0\text{ and }l_2x+m_2y+n_2=0\] Hence the equations of the lines through the origin parallel to the above lines are $l_1x+m_1y=0$ and $l_2x+m_2y=0$. Then, the combined equation is \[(l_1x+m_1y)(l_2x+m_2y)=0\] \[l_1l_2x^2+(l_1m_2+l_2m_1)xy+m_1m_2y^2=0\] \[\therefore ax^2+2hxy+by^2=0\]


Angle between two lines represented by \[ax^2+2hxy+by^2+2gx+2fy+c=0\text{ __(1)}\] is same as the angle between the two lines represented by \[ax^2+2hxy+by^2=0\text{ __(2)}\]

Since the two lines represented by the equation $\text{(2)}$ are parallel to the two lines represented by the equation $\text{(1)}$. So, the angle between the two lines given by $\text{(1)}$ is same as the angle between the lines given by $\text{(2)}$.

But the angle between the lines represented by $\text{(2)}$ is given by, \[\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}\] So, the angle between the two lines represented by equation $\text{(1)}$ is also given by the same formula \[\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}\] Hence, the two lines represented by \[ax^2+2hxy+by^2+2gx+2fy+c=0\] will be perpendicular to each other if $a+b=0$ and they will be parallel to each other is $h^2=ab$. [From: Angle between the line pair represented by $ax^2+2hxy+by^2=0$]


The point of intersection of the two lines represented by the general equation of second degree can be obtained by solving the separate equations \[l_1x+m_1y+n_1=0\text{ and }l_2x+m_2y+n_2=0\] for $x$ and $y$, and writing down the result in terms of the coefficients of the given equation by using the relations $l_1l_2=a$, $m_1m_2=b$, $n_1n_2=c$, $l_1m_2+m_1l_2=2h$, $l_1n_2+n_1l_2=2g$, $m_1n_2+m_2n_1=2f$. Then, the coordinates of the point of intersection will be \[\left(\sqrt{\frac{f^2-bc}{h^2-ab}},\sqrt{\frac{g^2-ca}{h^2-ab}}\right)\]


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