Question

X^2-2xy+y^2=4 |x+y|=2​

Answer 1

Ответ:

$x^{2} -2xy+y^{2}=4$

$|x+y|=2$

x+y=2 иди x+y=-2

Ответ: пара параллельных прямых у=2-х или у=-2-х

Объяснение:

Answer 2

The given equations are:

1. $x^2 - 2xy + y^2 = 4$
2. $|x + y| = 2$

Let's analyze these equations one by one:

Equation 1 represents a conic section, specifically an ellipse. It has the form of a general equation for an ellipse:

$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$

Comparing this general equation with the given equation $x^2 - 2xy + y^2 = 4$, we can see that $A = 1$, $B = -2$, $C = 1$, $D = 0$, $E = 0$, and $F = -4$.

The discriminant $B^2 - 4AC$ of the general equation determines the type of conic section. In this case, $B^2 - 4AC = (-2)^2 - 4(1)(1) = 4 - 4 = 0$, which means that the given equation represents a degenerate conic section, specifically a pair of intersecting lines. This is because the discriminant being zero indicates that the ellipse has been "squished" into two lines that intersect.

Equation 2 represents an absolute value equation involving $x + y$. The equation $|x + y| = 2$ means that the sum of $x$ and $y$ is either 2 or -2, because the absolute value of a number can be the number itself or its negation.

Now let's combine the two equations to see how they relate:

From equation 2, we have two cases:

1. $x + y = 2$
2. $x + y = -2$

Let's solve each case separately:

Case 1: $x + y = 2$ Solving this equation for $y$, we get $y = 2 - x$. Substituting this value of $y$ into equation 1, we get: $x^2 - 2x(2 - x) + (2 - x)^2 = 4$ Simplify and solve for $x$: $x^2 - 4x + 4 + x^2 - 4x + 4 = 4$ $2x^2 - 8x + 4 = 0$ $x^2 - 4x + 2 = 0$

This quadratic equation does not have rational solutions, so the solutions for $x$ will be complex.

Case 2: $x + y = -2$ Solving this equation for $y$, we get $y = -2 - x$. Substituting this value of $y$ into equation 1, we get: $x^2 - 2x(-2 - x) + (-2 - x)^2 = 4$ Simplify and solve for $x$: $x^2 + 2x^2 + 4x + 4 + x^2 + 4x + 4 = 4$ $4x^2 + 8x + 8 = 0$ $x^2 + 2x + 2 = 0$

This quadratic equation also does not have real solutions, so the solutions for $x$ will be complex.

In both cases, the solutions for $x$ are complex numbers, which means there are no real solutions that satisfy both equations. The intersection of the ellipse and the lines defined by $|x + y| = 2$ in the real plane is empty.