The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. Represents a second degree equation where a, h, b doesn’t variables simultaneously.
Let a ≠ 0.
Now, the above equation becomes
a2 x2 + 2ax (hy + g) = aby2 – 2afy – ac
on completing the square on the left side, we get,
a2 x2 + 2ax (hy + g) = y2 (h2 – ab) + 2y (gh – af) + g2 – ac.
i.e. (ax + hy + g) = + √y2(h2–ab)+2y(gh–af) +(g2–ac)
We cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign be a perfect square. The condition for this is,
(gh – af)2 = (h2 – ab) (g2 – ac)
hope it clears.
Now if Delta is 0
- h^2 – ab > 0 intersecting real lines
- h^2 – ab = 0. parallel lines
- h^2 – ab