Show that a homogeneous equations of degree two in x and y , i.e., `ax^(2) + 2 hxy + by^(2) = 0` represents a pair of lines passing through the origin if `h^(2) - 2ab ge 0`.

Explanation

Proof : Consider a homogeneous equation of the second degree in x and y,
`ax^(2) + 2hxy + by^(2) = 0" "…(1)`
Case I : If `b = 0` (i.e., `a ne 0, h ne = 0`), then the equation (1) reduces to `ax^(2) + 2hxy = 0` i.e., `x(ax + 2hy) = 0`
This represents the two lines `x = 0` and `ax + 2hy = 0`, both passing through the origin.
Case II : If a = 0 and b = 0 (i.e., `h ne 0`), then the equation (1) reduces to 2hxy = 0, i.e., xy = 0 which represents the coordinates axes and they pass through the origin.
Case III : if `b ne 0`, then the equation (1), on dividing it by b, becomes
`(a)/(b)x^(2) + (2hxy)/(b) +y^(2)=0 " "therefore y^(2) + (2h)/(b)xy=-(a)/(b)x^(2)`
On completing the square and adjusting, we get,
`y^(2) + (2h)/(b) xy + (h^(2)x^(2))/(b^(2))-(a)/(b)x^(2)`
` therefore (y + (h)/(b)x)^(2) = ((h^(2) -ab)/(b^(2)))x^(2)`
`therefore y + (h)/(b)x= pm(sqrt(h^(2) -ab))/(b)x`
`therefore y = (-h)/(b) x pm(sqrt(h^(2) -ab))/(b)x" "therefore y = ((-hpmsqrt(h^(2)-ab))/(b))x`
`therefore` the equation represents the two lines
``y = ((-h + sqrt(h^(2)-ab))/(b))x and y = ((-h-sqrt(h^(2) -ab))/(b))x`
Since none of these equations contains a constant term, both these lines pass through the origin.
THus homogenous equation (1) represents a pair of lines through the origin. if `h^(2) -ab ge 0`.