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noname [10]
3 years ago
5

MEDAL AND POINTS!!!!! write equation: The hyperbola with center (1, 2), vertices (1, 5) and (1, –1) , and foci (1, 7) and (1, –3

).
Mathematics
1 answer:
telo118 [61]3 years ago
8 0
We notice the x of the foci don't change
that means the hyperbola opens up and down
\frac{(y-k)^2}{a^2}-  \frac{(x-h)^2}{b^2}=1
hmm
center is (1,2) and vertices (1,5) and (1,-1)
the distance to a vertex is 3, that is 'a'
the rectangle of refernce is 6 units high
remember that
a²+b²=c² where a is major axis, b is minor axis and c is the distance from focs to center
from focus to center is 5 units
a is 3
3²+b²=5²
9+b²=25
mnus 9
b²=16
sqrt
b=4


so
\frac{(y-2)^2}{3^2}-  \frac{(x-1)^2}{4^2}=1 is the equation
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Answer:

15x - 4y =-50 -----(i)

3x - 2y=-16 --------(ii)

Using Substitution

From (i)

15x=4y-50

x=4y-50/15

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3x - 2y= -16

3[4y-50]/15 - 2y =-16

Multiply through by 15 to clear the fraction

3(4y - 50) - 30y=-240

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Substitute into any eqn of choice to get x

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3 years ago
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Olenka [21]

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5 0
4 years ago
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=====================================================

Explanation:

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