Question

how to find the double point of the curve y^2 = (x-2)^2 (x-1) ?

Answer 1

Answer on Question #70926 – Math – Calculus

Question

How to find the double point of the curve $y^2 = (x - 2)^2 (x - 1)$?

Solution

The function $f(x, y) = 0$ has a singular point $P$ if in point $P$ both first partial derivatives of the function are zero:

$$\frac{\partial}{\partial x} f(x, y) = \frac{\partial}{\partial y} f(x, y) = 0.$$

Here

$$f(x, y) = y^2 - (x - 2)^2 \cdot (x - 1) = y^2 - x^3 + 5x^2 - 8x + 4$$ $$\frac{\partial}{\partial x} f(x, y) = - (x - 2)^2 - (x - 1)(2x - 4) = -3x^2 + 10x - 8$$ $$\frac{\partial}{\partial y} f(x, y) = 2y$$ $$\frac{\partial^2}{\partial x^2} f(x, y) = 10 - 6x$$ $$\frac{\partial^2}{\partial y^2} f(x, y) = 2$$ $$\frac{\partial^2}{\partial x \partial y} f(x, y) = 0$$

To find a singular point, one should solve the system of equations

\left\{ \begin{array}{l} \frac{\partial}{\partial x} f(x, y) = 0 \\ \frac{\partial}{\partial y} f(x, y) = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} - (x - 2)^2 - (x - 1)(2x - 4) = 0 \\ 2y = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \left[ \begin{array}{c} x = \frac{4}{3} \\ x = 2 \\ y = 0 \end{array} \right] \Rightarrow \left[ \begin{array}{c} x = \frac{4}{3} \\ y = 0 \\ x = 2 \\ y = 0 \end{array} \right] \right.

Point $\left(\frac{4}{3}, 0\right)$ does not lie on the curve $y^2 = (x - 2)^2 (x - 1)$ because $f\left(\frac{4}{3}, 0\right) \neq 0$.

Point $(2,0)$ lies on the curve $y^2 = (x - 2)^2 (x - 1)$ because $f(2,0) = 0$.

Point $(2,0)$ is a double point because at least one of the second derivatives $\frac{\partial^2}{\partial x^2} f(x,y) \neq 0$, $\frac{\partial^2}{\partial y^2} f(x,y) \neq 0$ simultaneously for this point.

Type of the double point is determined by the expression

$$g(x, y) = \frac{\partial^2}{\partial x^2} f(x, y) \cdot \frac{\partial^2}{\partial y^2} f(x, y) - \left(\frac{\partial^2}{\partial x \partial y} f(x, y)\right)^2 = 2(10 - 6x) - 0^2 = 20 - 12x;$$ $$g(2, 0) = 20 - 12 \times 2 < 0.$$

In this case, there are two branches of the curve passing through the point $(2, 0)$. These branches have distinct tangents.

A slope $m$ of the tangent can be found from the equation

$$\frac{\partial^2 f}{\partial y^2}(x_0, y_0) m^2 + 2 \frac{\partial^2 f}{\partial x \partial y}(x_0, y_0) m + \frac{\partial^2 f}{\partial x^2}(x_0, y_0) = 0$$ $$2m^2 - 2 = 0$$ $$m = 1 \text{ or } m = -1$$

Curve $y^2 = (x - 2)^2(x - 1)$ can be rewritten as $y = (x - 2)\sqrt{x - 1}$ or

$$y = -(x - 2)\sqrt{x - 1}.$$

Such a point $(2, 0)$ is called a node (a point of self-intersection).

**Answer**: $(2, 0)$ is a double point, a point of self-intersection.

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