Pair of Straight Lines
8.0 Homogenization
8.0 Homogenization
This method is used to write the joint equation of two lines connecting origin to the points of intersection of a given line
$lx+my+n=0$ and a given second degree curve $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$. The equation is obtained by homogenizing the equation of curve with the help of equation of line i.e., $$a{x^2} + 2hxy + b{y^2} + 2gx\left( {\frac{{lx + my}}{{ - n}}} \right) + 2fy\left( {\frac{{lx + my}}{{ - n}}} \right) + c{\left( {\frac{{lx + my}}{{ - n}}} \right)^2} = 0$$
Note: Here we have written $1$ as $\frac{{lx + my}}{{ - n}}$ and converted all the terms of the curve to second-degree expression.
The above equation on simplification will be of the form $$A{x^2} + 2Hxy + B{y^2} = 0$$ and will represent the required straight lines.
If $\theta $ be the angle between them, then $$\theta = {\tan ^{ - 1}}\left( {\frac{{2\sqrt {({H^2} - AB)} }}{{\left| {A + B} \right|}}} \right)$$
Question 4. Prove that the angle between the lines joining the origin to the points of intersection of the straight line $y-3x=2$ with the curve ${x^2} + 2xy + 3{y^2} + 4x + 8y - 11 = 0$ is $${\tan ^{ - 1}}\frac{{2\sqrt 2 }}{3}$$
Solution: Equation of given curve is $${x^2} + 2xy + 3{y^2} + 4x + 8y - 11 = 0$$ and the given line is $$\begin{equation} \begin{aligned} y - 3x = 2 \\ \frac{{y - 3x}}{2} = 1 \\\end{aligned} \end{equation} $$
Making the given equation of curve a homogeneous equation of second degree in $x$ and $y$ with the help of given straight line i.e., $$\begin{equation} \begin{aligned} {x^2} + 2xy + 3{y^2} + 4x\left( {\frac{{y - 3x}}{2}} \right) + 8y\left( {\frac{{y - 3x}}{2}} \right) - 11{\left( {\frac{{y - 3x}}{2}} \right)^2} = 0 \\ {x^2} + 2xy + 3{y^2} + \frac{1}{2}(4xy + 8{y^2} - 12{x^2} - 24xy) - \frac{{11}}{4}({y^2} - 6xy + 9{x^2}) = 0 \\ - 119{x^2} + 34xy + 17{y^2} = 0 \\ 7{x^2} - 2xy - {y^2} = 0 \\\end{aligned} \end{equation} $$
which is the equation of lines joining the origin to the points of intersection of straight line and curve. Compare it with the equation $$a{x^2} + 2hxy + b{y^2} = 0$$ we get,
$a=7,\ b=-1,\ and\ h=-1$.
If $\theta $ is the acute angle between the pair of lines, then
$$\begin{equation} \begin{aligned} \theta = {\tan ^{ - 1}}\left( {\frac{{2\sqrt {({h^2} - ab)} }}{{\left| {a + b} \right|}}} \right) \\ {\text{ = }}{\tan ^{ - 1}}\left( {\frac{{2\sqrt {(1 + 7)} }}{{\left| {7 - 1} \right|}}} \right) \\ {\text{ = }}{\tan ^{ - 1}}\left( {\frac{{2\sqrt 2 }}{3}} \right) \\\end{aligned} \end{equation} $$
