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Sagot :
Given:
Hyperbola with directrices at y = ±2 and foci at (0, 6) and (0, −6).
To find:
The equation of hyperbola.
Solution:
We have, directrices at y = ±2 so this hyparabola is along the y-axis.
The standard form of hyperbola is
[tex]\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1[/tex] ...(i)
where, (h,k) is center, foci are [tex](h,k\pm c)[/tex] and directrix are [tex]y=k\pm \dfrac{a^2}{c}[/tex].
On comparing foci, we get
[tex](h,k\pm c)=(0,\pm 6)[/tex]
[tex]h=0,k=0,c=6[/tex]
On comparing directrix we get
[tex]k\pm \dfrac{a^2}{c}=\pm 2[/tex]
[tex]\dfrac{a^2}{c}=2[/tex]
[tex]\dfrac{a^2}{6}=2[/tex]
[tex]a^2=12[/tex]
Now,
[tex]a^2+b^2=c^2[/tex]
[tex]12+b^2=(6)^2[/tex]
[tex]b^2=36-12[/tex]
[tex]b^2=24[/tex]
Putting [tex]h=0,k=0,a^2=12,b^2=24[/tex], we get
[tex]\dfrac{(y-0)^2}{12}-\dfrac{(x-0)^2}{24}=1[/tex]
[tex]\dfrac{y^2}{12}-\dfrac{x^2}{24}=1[/tex]
Therefore, the equation of hyperbola is [tex]\dfrac{y^2}{12}-\dfrac{x^2}{24}=1[/tex].
Answer:
the person above has the right answer. I took the test.
Step-by-step explanation:
The person above is correct.
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