Homogeneous Equations in Two Variables - Practice Questions & MCQ

Homogeneous Equations in Two Variables

Homogeneous equations are those equations where each term has the same degree of variable terms.
The equation $a x^2+2 h x y+b y^2=0$ is a homogeneous equation of second degree.
It always represents two straight lines passing through the origin.
The given equation is :

$
a x^2+2 h x y+b y^2=0
$
Divide the equation by $x^2$

$
\Rightarrow \quad a+2 h\left(\frac{y}{x}\right)+b\left(\frac{y^2}{x^2}\right)=0
$

now put $\frac{y}{x}=m$

$
\Rightarrow \quad \mathrm{a}+2 \mathrm{~h}(\mathrm{~m})+\mathrm{b}\left(\mathrm{~m}^2\right)=0 \quad \ldots \text { (ii) }
$
If $\mathrm{m}_1$ and $\mathrm{m}_2$ are roots, then

$
\mathrm{m}_1+\mathrm{m}_2=-\frac{2 \mathrm{~h}}{\mathrm{~b}}
$

amd, $\quad \mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{b}}{\mathrm{a}}$
Thus, $\mathrm{y}=\mathrm{m}_1 \mathrm{x}$ and $\mathrm{y}=\mathrm{m}_2 \mathrm{x}$ are two straight lines given by Eq. (i)

Real and Imaginary Lines

For equation (ii), if roots $m_1$ and $m_2$ are real, then the lines are also real, and if $m_1$ and $m_2$ are not real (imaginary), then the lines are also not real (imaginary). This depends on discriminant of equation (ii)
Discriminant $=4 \mathrm{~h}^2-4 \mathrm{ab}=4\left(\mathrm{~h}^2-\mathrm{ab}\right)$
1. The lines are real and distinct if $D>0: h^2-a b>0$
2. The lines are coincident if $D=0: h^2-a b=0$
3. The lines are imaginary if $D<0: h^2-a b<0$

Angle between pair of Straight Lines
General equation of pair of straight line is $\mathrm{ax}^2+2 \mathrm{hxy}+\mathrm{by}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$. Angle between the pair of the straight lines is

$
\theta=\tan ^{-1}\left\{\frac{2 \sqrt{\mathrm{~h}^2-\mathrm{ab}}}{|\mathrm{a}+\mathrm{b}|}\right\}
$
 

Homogenization of Second Degree Equation

Consider the general second degree equation

$
S \equiv a x^2+b y^2+2 h x y+2 g x+2 f y+c=0
$
This equation may represent a pair of the straight lines, circle, parabola, ellipse or hyperbola.

Now, let's a straight line : $\mathrm{L}=\mathrm{lx}+\mathrm{my}+\mathrm{n}=0$
(2) intersects the curve at two points $A$ and $B$.

The combined equation of pair of straight lines OA and OB must be homogenous and must contain only second-degree terms.
In order to make Eq (1) homogeneous using Eq (2), we can use (2) to writte

$
1=\frac{\mathrm{lx}+\mathrm{my}}{-\mathrm{n}}
$

now, equation of curve is

$
S \equiv\left(a x^2+2 h x y+b y^2\right)+(2 g x+2 f y) \cdot 1+c \cdot 1=0
$

put the value of 1 in above equation, we get

$
\left(a x^2+2 h x y+b y^2\right)+(2 g x+2 f y)\left(\frac{l x+m y}{-n}\right)+c\left(\frac{l x+m y}{-n}\right)^2=0
$
This will give a homogeneous equation of degree 2, and this represents the combined equation of OA and OB.