What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is,

Question

(x+3)^2 - 4(y-2)^2 = 4?

Answer 1

The hyperbola equation is (x + 3)² - 4(y - 2)² = 4.

The standard form of the equation of a hyperbola with center (h, k) (where a and b are not equals to 0) is (x - h)²/a² - (y - k)²/b² = 1 (Transverse axis is horizontal) or (y - k)²/a² - (x - h)²/b² = 1 (Transverse axis is vertical).

The vertices and foci are, respectively a and c units from the center (h, k)  and the relation between a, b and c is b² = c² - a².

Write the equation (x + 3)² - 4(y - 2)² = 4 in the standard form of equation of hyperbola.

(x + 3)²/4 - 4(y - 2)²/4 = 4/4

(x + 3)²/4 - (y - 2)²/1 = 1.

Compare the equation (x + 3)²/4 - (y - 2)²/1 = 1 with (x - h)²/a² - (y - k)²/b² = 1.

a² = 4, b² = 1, k = 2 and h = - 3.

a = ± 2 and b = ± 1.

To find the value of c, substitute the value of a² = 4 and b² = 1 in b² = c² - a².

1 = c² - 4

5 = c²

c = ± √5.

Here the transverse axis is horizontal, the asymptotes are of the forms y = (b/a) x and y = - (b/a) x.

The asymptotes equations are y = ± 1/2 x.

The slopes of asymptotes are ± 1/2,

Center = (h, k) = (- 3, 2),

Vertices = (h ± a, k) = (- 3 ± 9, 2) = (6, 2) and (- 12, 2),

Foci = (h ± c, k) = (- 3 ± √5, 0) = (- 3 + √5, 0) and (- 3 - √5, 0).

 

Comments

typo-----mistake. Foci = (h ± c, k) = (- 3 ± √5, 2) = (- 3 + √5, 2) and (- 3 - √5, 2)

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Vertices (h ± a, k)

In this case (h ,k ) = (-3, 2) and a = ± 2 .

Substitute the above values (- 3 ± 2, 2) 

= (- 3 + 2, 2) (- 3 - 2, 2)

 Vertices : (-1, 2)(-5, 2).