Question

Identify the conic x^2- 2xy + y^2 + √2x = 2

Answer 1

Let's solve this problem

first of all we simplify expression

$x^2 - 2 x y + y^2 + \sqrt{2 x} = 2$

$x^2 - 2 x y + \sqrt{2} \sqrt{x} + y^2 = 2$

The graph of conic

Slutions:

Solve for y:

$\sqrt{2}\sqrt{x}+x^2-2xy+y^2 = 2$

we solve quadratic equation by completing the square.

Subtract $\sqrt{2}\sqrt{x}+x^2$ from the both sides.

$y^2-2xy = 2-\sqrt{2}\sqrt{x}-x^2$

$D = b^2-4ac = 4x^2+4(2-\sqrt2\sqrt{x}-x^2) = 8 - 4\sqrt{2}\sqrt{x}$

$x_{1,2} = \large\frac{-b\pm \sqrt{D}}{2a}$ = $\large\frac{2x\pm\sqrt{8-4\sqrt{2x}}}{2}$ $= x \pm \sqrt{2 - \sqrt{2x}}$

Integer solutions: x = 2, y = 2

Implicit derivtives:

$\large\frac{\delta x(y)}{dy} = \frac{4 x - 2 x^3 - 4 y + 6 x^2 y - 6 x y^2 + 2 y^3}{1 + 4 x - 2 x^3 - 4 y + 6 x^2 y - 6 x y^2 + 2 y^3}$

$\large\frac{dy}{\delta x(y)} = \frac{1 + 4 x - 2 x^3 - 4 y + 6 x^2 y - 6 x y^2 + 2 y^3}{4 x - 2 x^3 - 4 y + 6 x^2 y - 6 x y^2 + 2 y^3}$