Question

Show that every homogeneous equation of degree two in x and y, i.e., ax² + 2hxy + by² = 0 represents a pair of lines passing through origin if h²−ab≥0.

Answer 1

Consider a homogeneous equation of the second degree in x and y,

${\displaystyle a{x}^2+2hxy+b{y}^2=0......................\left(1\right)}$

Case I: If b = 0 (i.e., a ≠ 0, h ≠ 0 ), then the equation (1) reduce to ax²+ 2hxy= 0
i.e., x(ax + 2hy) = 0

Case II: If a = 0 and b = 0 (i.e. h ≠ 0 ), then the equation (1) reduces to 2hxy = 0, i.e., xy = 0 which represents the coordinate axes and they pass through the origin.

Case III: If b ≠ 0, then the equation (1), on dividing it by b, becomes ${\displaystyle \frac{a}{b}{x}^2+\frac{2hxy}{b}+y^2=0}$

${\displaystyle \therefore y^2+\frac{2h}{b}xy=-\frac{a}{b}{x}^2}$

On completing the square and adjusting, we get ${\displaystyle y^2+\frac{2h}{b}xy+\frac{h^2{x}^2}{b^2}=\frac{h^2{x}^2}{b^2}-\frac{a}{b}{x}^2}$

${\displaystyle {(y+\frac{h}{b}x)}^2=\left(\frac{h^2-ab}{b^2}\right)x^2}$

${\displaystyle \therefore y+\frac{h}{b}x=\pm \frac{\sqrt{h^2-ab}}{b}x}$

${\displaystyle \therefore y=-\frac{h}{b}x\pm \frac{\sqrt{h^2-ab}}{b}x}$

${\displaystyle \therefore y=\left(\frac{-h\pm \sqrt{h^2-ab}}{b}\right)x}$

∴ equation represents the two lines ${\displaystyle y=\left(\frac{-h+\sqrt{h^2-ab}}{b}\right)x\text{and}y=\left(\frac{-h-\sqrt{h^2-ab}}{b}\right)x}$

The above equation are in the form of y = mx
These lines passing through the origin.
Thus the homogeneous equation (1) represents a pair of lines through the origin, if h²- ab ≥ 0.

Answer 2

Consider a homogeneous equation of degree two in x and y

${\displaystyle a{x}^2+2hxy+b{y}^2=0.......................\left(i\right)}$

In this equation at least one of the coefficients a, b or h is non zero. We consider two cases.

Case I: If b = 0 then the equqtion

${\displaystyle a{x}^2+2hxy=0}$

${\displaystyle x(ax+2hy)=0}$

This is the joint equation of lines x = 0 and (ax+2hy)=0
These lines pass through the origin.

Case II: If b ≠ 0
Multiplying both the sides of equation (i) by b, we get

${\displaystyle ab{x}^2+2hbxy+b^2{y}^2=0}$

${\displaystyle 2hbxy+b^2{y}^2=-ab{x}^2}$

To make LHS a complete square, we add h²x² on both the sides.

${\displaystyle b^2{y}^2+2hbxy+h^2{y}^2=-ab{x}^2+h^2{x}^2}$

${\displaystyle {(by+hx)}^2=(h^2-ab)x^2}$

${\displaystyle {(by+hx)}^2={\left[\left(\sqrt{h^2-ab}\right)x\right]}^2}$

${\displaystyle {(by+hx)}^2-{\left[\left(\sqrt{h^2-ab}\right)x\right]}^2=0}$

${\displaystyle [(by+hx)+\left[\left(\sqrt{h^2-ab}\right)x\right]][(by+hx)-\left[\left(\sqrt{h^2-ab}\right)x\right]]=0}$

It is the joint equation of two lines

${\displaystyle (by+hx)+[\left(\sqrt{h^2-ab}\right)x=0\text{and}(by+hx)-[\left(\sqrt{h^2-ab}\right)x=0}$

${\displaystyle (h+\sqrt{h^2-ab})x+by=0\text{and}(h-\sqrt{h^2-ab})x+by=0}$

These lines pass through the origin when h²-ab>0

From the above two cases we conclude that the equation ${\displaystyle a{x}^2+2hxy+b{y}^2=0}$ represents a pair of lines passing through the origin.