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Which is the equation of a hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0)? y squared over 40 minus x squared over 10 equals 1 y squared over 10 minus x squared over 15 equals 1 x squared over 10 minus y squared over 40 equals 1 x squared over 10 minus y squared over 15 equals 1

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MrRoyalidn

The equation of the hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0) is [tex]\frac{x^2}{10} + \frac{y^2}{15} = 1[/tex]

How to determine the equation of the hyperbola?

The given parameters are:

  • Directrices at x = ±2
  • Foci at (5, 0) and (−5, 0)

The foci of a hyperbola are represented as:

Foci = (k ± c, h)

The center is:

Center = (h,k)

And the directrix is:

Directrix, x = h ± a²/c

By comparison, we have:

k ± c = ±5

h = 0

h ± a²/c = ±2

Substitute h = 0 in h ± a²/c = ±2

0 ± a²/c = ±2

This gives

a²/c = 2

Multiply both sides by c

a² = 2c

k ± c = ±5 means that:

k ± c = 0 ± 5

By comparison, we have:

k = 0 and c = 5

Substitute c = 5 in a² = 2c

a² = 2 * 5

a² = 10

Next, we calculate b using:

b² = c² - a²

This gives

b² = 5² - 10

Evaluate

b² = 15

The hyperbola is represented as:

[tex]\frac{(x - k)^2}{a^2} + \frac{(y - h)^2}{b^2} = 1[/tex]

So, we have:

[tex]\frac{(x - 0)^2}{10} + \frac{(y - 0)^2}{15} = 1[/tex]

Evaluate

[tex]\frac{x^2}{10} + \frac{y^2}{15} = 1[/tex]

Hence, the equation of the hyperbola is [tex]\frac{x^2}{10} + \frac{y^2}{15} = 1[/tex]

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