# Thread: Finding singular points of a curve in affine space

1. ## Finding singular points of a curve in affine space

How do i find the singular points of the following curve in A^2(C) OR prove that there are no singular points? (where A is affine space and C is complex numbers)

The curve is

(x^2)(y^2)-6(x^2)+2xy-6(y^2)+25=0

Taking partial derivatives i get the jacobian matrix

J = (2x(y^2)-12x+2y 2(x^2)y+2x-12y)

I do not know how to proceed from here.

Any help is appreciated.

Thanks

2. Originally Posted by Siknature
How do i find the singular points of the following curve in A^2(C) OR prove that there are no singular points? (where A is affine space and C is complex numbers)

The curve is

$\displaystyle x^2y^2-6x^2+2xy-6y^2+25=0$

Taking partial derivatives i get the jacobian matrix

$\displaystyle J = \begin{pmatrix}2xy^2-12x+2y & 2x^2y+2x-12y\end{pmatrix}$
I do not know how to proceed from here.
The condition for a singular point is that the Jacobian should have less than full rank. The full rank of a 1×2 matrix is 1. So at a singular point, the matrix must have rank 0, which is the same as saying that both entries should be 0.

Therefore you need to solve the equations
\displaystyle \left.\begin{aligned}2xy^2-12x+2y &= 0, \\ 2x^2y+2x-12y &= 0.\end{aligned}\right\}
An efficient way to do that is to add the equations, getting $\displaystyle (xy-5)(x+y)=0$. So either $\displaystyle xy=5$, leading to the solutions $\displaystyle x=y=\pm\sqrt5$; or $\displaystyle x+y=0$, leading to the solutions $\displaystyle x=-y=\pm\sqrt7$ and also the solution $\displaystyle x=y=0$.

(You then have to check whether any of those points actually lie on the curve. If they don't, they obviously don't yield singular points.)

### how to find singular points of the curve o

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