Precalculus help?

Question

Write the equations of the asymptotes and foci
-2x^2 + y^2 + 4x + 6y= -3

Answer 1

The conic equation - 2x² + y² + 4x + 6y = - 3

The coefficients of x² and y² are of opposite sign, then it will be hyperbola.

The above equation is represents a hyperbola.

Write the equation - 2x² + y² + 4x + 6y = - 3 in standard form of hyperbola.

- 2x² + y² + 4x + 6y = - 3

- 2x² + 4x + y² + 6y = - 3

- 2(x² - 2x) + 1(y² + 6y) = - 3

To change the expression into a perfect square  add (half of the x coefficient)² and (half of the y - coefficient)² to each side of the expression.

Here x coefficient = - 2, so, (half the x coefficient)² = (- 2/2)² = 1.

Here y coefficient = 6, so, (half the x coefficient)² = (6/2)²= 9.

Add - 2 and 9 to each side of the equation.

- 2(x² - 2x) + 1(y² + 6y) - 2 + 9 = - 3 - 2 + 9

- 2(x² - 2x + 1) + 1(y² + 6y + 9) = 4

[- 2(x - 1)²]/4 + (y + 3)²/4 = 4/4

(x - 1)²/(-2) + [y - (- 3)]²/4 = 1

[y - (- 3)]²/2² - (x - 1)²/(√2)² = 1

Compare it to standard form of vertical hyperbola (y - k)²/a² - (x - h)²/b² = 1

Where a = 2, b = √2

Center (h, k) = ( 1, - 3)

Foci (h, k + c) (h , k - c)

c = √(a² + b²)

c = √[2² + (√2)²] = √(4 + 2)

c = √6

Foci = (1, -3+√6) , (1, -3-√6).

And the equations for asymptotes are y = ± (a/b)(x - h) + k

y = ± (2/√2)(x - 1) + 3

y = ± (√2)(x - 1) + 3

y = ± (√2)(x - 1) - 3

y = + (√2)(x - 1) - 3 and y = - (√2)(x - 1) - 3

Asymptotes are y = √2x - √2 - 3 and y = - √2x + √2 - 3.