Question

X^2+4y^2+4x-16y-8=0​

Answer 1

Ответ:

(х+2у)^2+(2х-4у)=8.

Answer 2

The equation you've provided is a quadratic equation in two variables, x and y. It represents the equation of an ellipse in its standard form. To better understand and work with this equation, let's rearrange it:

x^2 + 4x + 4y^2 - 16y - 8 = 0

Complete the square for both the x and y terms:

(x^2 + 4x + 4) + (4y^2 - 16y + 16) - 8 - 4*16 = 0

(x + 2)^2 + 4(y - 2)^2 - 64 = 0

Now, the equation is in standard form for an ellipse:

(x + 2)^2 + 4(y - 2)^2 = 64

This is the equation of an ellipse centered at the point (-2, 2) with a horizontal major axis (along the x-axis) and a vertical minor axis (along the y-axis). The lengths of the major and minor axes are determined by the value of the constant on the right side of the equation. In this case, the major axis has a length of 16 (2 * √64) and the minor axis has a length of 8 (2 * √4).