Question

5х^2+9у^2-12ху-10х+25=0​

Answer 1

Ответ и объяснение: в приложении

Answer 2

The equation you've provided is in the form of a quadratic equation with two variables, x and y. It represents a conic section known as an ellipse. The general form of an ellipse equation is:

Ax^2 + By^2 + Cxy + Dx + Ey + F = 0

In your equation: A = 5 B = 9 C = -12 D = -10 E = 0 (since there's no y term without y) F = 25

So, the equation:

5x^2 + 9y^2 - 12xy - 10x + 25 = 0

represents an ellipse. The center, major and minor axes, and eccentricity of the ellipse can be determined by completing the square. However, before proceeding with that, let's first simplify the equation to its standard form by removing the linear term (10x) to prepare it for completing the square:

5x^2 + 9y^2 - 12xy + 25 = 0

Now, to find the center, we need to rewrite the equation in a perfect square format:

(5x^2 - 12xy) + 9y^2 + 25 = 0

Next, we complete the square for the x terms inside the parentheses:

5(x^2 - (12/5)xy) + 9y^2 + 25 = 0

To complete the square, we need to add and subtract the square of half of the coefficient of the xy term inside the parentheses:

5(x^2 - (12/5)xy + (36/25)y^2 - (36/25)y^2) + 9y^2 + 25 = 0

Now, we can factor the first three terms inside the parentheses as a perfect square trinomial:

5((x - (6/5)y)^2 - (36/25)y^2) + 9y^2 + 25 = 0

Expand the squared term and simplify:

5(x - (6/5)y)^2 - 36y^2 + 9y^2 + 25 = 0

Combine like terms:

5(x - (6/5)y)^2 - 27y^2 + 25 = 0

Now, to make the equation look like the standard form of an ellipse, we'll divide the entire equation by 25:

(x - (6/5)y)^2 / 5^2 - 27y^2 / 5^2 + 25 / 25 = 0

Simplify further:

(x - (6/5)y)^2 / 25 - 27y^2 / 25 + 1 = 0

Finally, the equation is in standard form:

(x - (6/5)y)^2 / 25 - 27y^2 / 25 + 1 = 0

The center of the ellipse is at the point (0, 0), the major axis is along the x-axis with a length of 10 units (2 * 5), the minor axis is along the y-axis with a length of 2 * sqrt(27) ≈ 10.39 units, and the eccentricity is the square root of (1 - (b^2 / a^2)), where a = 5 and b ≈ 10.39. Thus, the eccentricity is approximately sqrt(1 - (10.39^2 / 5^2)) ≈ 0.748.

Remember, this equation represents an ellipse, and these properties characterize its shape and orientation in the xy-plane.