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A hyperbola is centered at C = (-6,8) opens with vertices at (-6,11) and (-6,5), and...

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kristenmarieb... | Student, Grade 10 | (Level 1) Valedictorian

Posted July 6, 2013 at 4:58 PM via web

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A hyperbola is centered at C = (-6,8) opens with vertices at (-6,11) and (-6,5), and the slopes of the asymptotes are m = +- 3/7.  

The equation of the hyperbola is:

a) (x+6)^2 - (y-8)^2 = 1

or 

b) (y-8)^2 - (x+6)^2 = 1

Find the missing denominators. 

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llltkl | College Teacher | (Level 3) Valedictorian

Posted July 6, 2013 at 5:40 PM (Answer #1)

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The given hyperbola has its center at C = (-6, 8) and opens with vertices at (-6, 11) and (-6, 5) respectively. Since the center, and the vertices all have changing y-coordinate, the hyperbola is one with its transverse axis vertical.

Therefore, equation must have negative x-term, i.e. the one similar to the equation in option b).

The vertices of a hyperbola are placed ‘a’ distance away from its centre. With center at c=(-6, 8) and changing y-coordinates, the coordinates of its vertices would be at (-6, 8+-a).

Comparing its given vertices at (-6, 11) and (-6, 5), we get,

`8+a=11`

`rArr a=3`

and again, `8-a=5`

`rArr a=3` .

The slope of the asymptotes is given by `+-a/b` (transverse axis, vertical)

Comparing the given values yield `a/b=3/7` putting the value of a,

`rArr 3/b=3/7`

`rArr b=7`

Therefore, the required equation of the hyperbola is  

`(y-8)^2/3^2-(x+6)^2/7^2=1`

`rArr (y-8)^2/9-(x+6)^2/49=1`

Hence the denominators of the two terms in equation b) are 9 and 49 respectively.

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