# The general Equation of a conic section is ax^2+2hxy+by^2+2gx+2fy+c=0; Then how did we get the discriminant as abc+2fgh-af^2-bg^2-ch^2 ??

148 Points
14 years ago

Dear Moni

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)

i.e. g2h2 – 2afgh + a2f2 = g2h2 – abg2 – abg2 – ach2 + a2bc

cancelling and diving by a, we have the required condition

abc + 2fgh – af2 – af2 – bg2 – ch2 = 0

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