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coldgirl [10]
2 years ago
10

WILL MARK BRAINLIEST!! Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y = ±4 div

ided by 3.x. y squared over 9 minus x squared over 16 = 1 y squared over 36 minus x squared over 64 = 1 y squared over 64 minus x squared over 9 = 1 y squared over 64 minus x squared over 36 = 1
Mathematics
1 answer:
meriva2 years ago
6 0

Answer:

y^2 / 64 - x^2 /36 =1

Step-by-step explanation:

hyperbola with vertices at (0, ±8)

asymptotes at y = ±4/3x

The general form is

y^2 /a^2 - x^2/ b^2 =1

where the asymptote is y = ± a/b x

but we also have The vertex is ( 0 ,a)

rewriting the asymptote as 8/6 so a =8 and b = 6

y^2 / 8^2 -x^2/6^2 = 1

y^2 / 64 - x^2 /36 =1

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Answer:

On a coordinate plane, 2 quadrilaterals are shown. The larger quadrilateral has points (negative 5, 3), (1, 3), (4, 0), (negative 2, 0). The smaller quadrilateral has points (negative 1, 0), (negative 2, 1), (0, 1), and (1, 0).      

Step-by-step explanation:

<u>Larger quadrilateral</u>

Length of the top and bottom segments are 6 units.

Length of the left and right sides:

d = √[(-5 - (-2))² + (3 - 0)²] = √[3² + 3²] = √(2*3²) = 3√2 units

From point (-5, 3) to point (-2, 0) the slope is:

m = (3 - 0)/(-5 - (-2)) = -1

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Length of the top and bottom segments are 2 units.

Length of the left and right sides:

d = √[(-2 - (-1))² + (1 - 0)²] = √[1² + 1²] = √2 units

From point (-2, 1) to point (-1, 0) the slope is:

m = (1 - 0)/(-2 - (-1)) = -1

If you dilate the smaller quadrilateral by a factor of 3, you get the larger quadrilateral (they have the same slope, and 3 times length of the smaller quadrilateral is equal to the length of the larger quadrilateral).

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Answer:

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