Grade 11Analytical Geometry

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 ??

Moni RS

16 Years agoGrade 11

1 Answer

Badiuddin askIITians.ismu Expert

16 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

^{Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here
and we will get you the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Regards,
Askiitians Experts
Badiuddin}