Question

NEED HELP ASAP PLEASE WILL MARK BRANLIEST What would the equation of a hyperbola with directrices at x=±2 and foci at (6,0) (-6,0) be?

Answer 1

Answer:

The equation of the hyperbola can be presented as follows;

$\dfrac{\left x \right ^{2}}{12} - \dfrac{\left y \right ^{2}}{24} = 1$

Step-by-step explanation:

We have the equation of an hyperbola given as follows;

$\dfrac{\left (x - h \right )^{2}}{a^{2}} - \dfrac{\left (y - k \right )^{2}}{b^{2}} = 1$

The foal length = a² + b² = c²

c = focal length

The directrix = a²/c

The focal point are (h + c, k) and (h - c, k)

Therefore, by comparison with the given focal points, (6, 0) and (-6, 0), we have;

k = 0 and h + c = 6, h - c = -6

Therefore;

6 - c - c = -6

-2·c = -12

c = 6

h = 0

a²/c = 2

a² = 2 × 6 = 12

a = 2·√3

12 + b² = 6²

b² = 6² - 12 = 24

b² = 24

b = 2·√6

The equation of the hyper bola can then be written as follows;

$\dfrac{\left (x - 0 \right )^{2}}{12} - \dfrac{\left (y - 0 \right )^{2}}{24} = 1$

Which gives

$\dfrac{\left x \right ^{2}}{12} - \dfrac{\left y \right ^{2}}{24} = 1$

x² - y² = 12

y² = x² - 12.