Question

5x^2 - y^2 +10y-15=0​

Answer 1

Ответ:

(a, 5 + √5a), (a, 5 - √5a) a ∈ R, бесконечно много решений

Пошаговое объяснение:

5x² - (y² - 10y + 25) + 10 = 0

5x² - (y - 5)² = -10

(y - 5 - √5x)(y - 5 + √5x) = 0

y = 5 + √5x, x∈R

y = 5 - √5x, x∈R

Answer 2

The given equation is 5x^2 - y^2 + 10y - 15 = 0.

To analyze this equation, we can first try to rearrange it in a more manageable form. Let's group the terms involving y together:

5x^2 - (y^2 - 10y) - 15 = 0.

To complete the square for the terms involving y, we need to add and subtract the square of half the coefficient of y (in this case, 10/2 = 5):

5x^2 - (y^2 - 10y + 25 - 25) - 15 = 0.

Now, we can rewrite the equation as:

5x^2 - (y - 5)^2 + 25 - 15 = 0.

Simplifying further:

5x^2 - (y - 5)^2 + 10 = 0.

Now, we have the equation in a standard form, where we can identify the coefficients. From this form, we can see that the equation represents a hyperbola. Specifically, the equation represents a hyperbola centered at (0, 5) with a horizontal transverse axis.

The general form of a hyperbola centered at (h, k) is:

(x - h)^2/a^2 - (y - k)^2/b^2 = 1,

where a and b represent the lengths of the transverse and conjugate axes, respectively.

Comparing the given equation to the general form, we can identify the values:

Center: (h, k) = (0, 5), Transverse axis length: 2a = √10 (since a^2 = 10/5 = 2), Conjugate axis length: 2b = √10 (since b^2 = 10/5 = 2).

Therefore, the equation represents a hyperbola centered at (0, 5) with a transverse axis length of √10 and a conjugate axis length of √10.