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Sagot :
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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