I need help please with this
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The equation of the hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0) is
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:
So, we have:
Evaluate
Hence, the equation of the hyperbola is
Read more about hyperbola at:
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Answer:
Step-by-step explanation:
Cosine is the adjacent side over the hypotenuse (You can remember sin, cos, and tan by using sohcahtoa or sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite over adjacent). I think a picture would help, too.
I attached a picture of what I think the triangle would look like.
If the picture is right (we're assuming it is) and going with what we're given (the triangle was addressed as triangle XYZ, meaning that <u>angle Y is in the middle and that's the one we'll use</u>).
Looking at my picture then:
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