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

Question

y^2 - 4x^2 - 2y - 16x + 1 = 0?

Answer 1

The hyperbola equation is y² - 4x² - 2y -16x + 1 = 0.

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 y² - 4x² - 2y -16x + 1 = 0 in the standard form of equation of hyperbola.

- y² + 4x² + 2y +16x = 1.

(4x² + 16x) - (y² - 2y) = 1.

4(x² + 4x) - (y² - 2y) = 1.

To change the expressions (x² + 4x) and (y² - 2y) into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

 Here x coefficient = 4. so, (half the x coefficient)² = (4/2)²= 4.

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

Add 4(4) = 16 and subtract 1 from each side.

4(x² + 4x + 4) - (y² - 2y + 1) = 1 + 16 - 1.

4(x + 2)² - (y - 1)² = 16

(x + 2)²/4 - (y - 1)²/16 = 1.

(x + 2)²/2² - (y - 1)²/4² = 1

Compare the equation (x + 2)²/2² - (y - 1)²/4² = 1 with (x - h)²/a² - (y - k)²/b² = 1.

a = 2, b = 4, h = - 2 and k = 1.

To find the value of c, substitute the value of a² = 4 and b² = 16 in b² = c² - a².

16 = c² - 4

20 = c²

c = ± 2√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 = ± 4/2 x -------> y = ± 2x.

The slopes of asymptotes are ± 2,

Center = (h, k) = (- 2, 1),

Vertices = (h ± a, k) = (- 2 ± 2, 1) = (0, 1) and (- 4, 1),

Foci = (h ± c, k) = (- 2 ± 2√5, 1) = (- 2 + 2√5, 1) and (- 2 - 2√5, 1).

 

Comments

In the above answer, asymptotes are wrong.

Here the transverse axis is horizontal, so the asymptotes form of y - k = ± (b/a) (x - h).

h = - 2, k = 1, b = 4 and a = 2.

The asymptotes are y - (1) = (4)/(2) [ x - (- 2) ] and y - (1) = - (4)/(2) [ x - (- 2) ].

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

y - 1 = 2x + 4 and y - 1 = - 2x - 4

y = 2x + 5 and y = - 2x - 3.

The asymptotes are y = 2x + 5 and y = - 2x - 3.

The slopes of asymptotes = ± (b/a) = ± 4/2 = ± 2.

The slopes of asymptotes are ± 2.