Question

Good morning friends! :)
I need help with this equation

[( ((xo-1)*(y-1))/(yo-1) + 1)^2 + 2y^2 +3((zo-1)(y-1))/(yo-1) + 1)]^2 =1.

Thank you for sugestions!

Answer 1

I need to fint decriminant

Answer 2

You may find the discriminant using
B^2 -4ac

Answer 3

Result is
D=4[2((xo-1)/(yo-1))^2 -3((xo-1)*(zo-1))/(yo-1)*(yo-1) +3((zo-1)/(yo-1))^2 -2(xo-1)/(yo-1) -6(zo-1)/(yo-1)+3]=0

How to get it.
Yes i am using b^2-4ac

Answer 4

This is super confusing, the discriminant is something in a quadratic formula, this is not quadratic. Is there another definition?

Answer 5

This is question about matrix?

Answer 6

It will be more easy if you upload the picture :D

Answer 7

\([( ((xo-1)*(y-1))/(yo-1) + 1)^2 + 2y^2 +3((zo-1)(y-1))/(yo-1) + 1)]^2 =1\)
There are mismatched parentheses in the above expression.
Is it:
\(\Large ((\frac{(xo-1)*(y-1)}{yo-1}+1)^2+2y^2 +\frac{3(zo-1)*(y-1)}{yo-1}+1)^2 =1\)
if not, please edit/correct.

Please, in the future, double-check complicated expressions before/after posting. Helper's who spend time decoding your expression cannot help other users.
Also, if possible, write in LaTeX format so you can do a good verification.

Answer 8

#Matmate It is 3*[(zo-1)*(y-1)/(yo-1) +1]^2 ,my mistake.

Answer 9

Rest of expression is correct.

Answer 10

Like this?
\(\Large (\frac{(xo-1)*(y-1)}{yo-1}+1)^2+2y^2 +3(\frac{(zo-1)*(y-1)}{yo-1}+1)^2 =1\)
It makes a little more sense, because now it is a quadratic in y.

Answer 11

To help you expand the expression a little easier, substitute
x1=x0-1, y1=y0-1,z1=z0-1.

Answer 12

Yes it is like you wrote.Alright I'll substitute that.