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Given the equation `3x^2 - 12y^2 + 6x + 48y - 93 = 0` determine: a)  The Center b)...

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted July 6, 2013 at 2:24 PM via web

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Given the equation `3x^2 - 12y^2 + 6x + 48y - 93 = 0` determine:

a)  The Center

b)  The 2 Vertices

c)  The slopes of the asymptotes (enter as a reduced number)

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mjripalda | High School Teacher | (Level 3) Educator

Posted July 6, 2013 at 3:12 PM (Answer #1)

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`3x^2-12y^2+6x+48y-93=0`

(a) Center

Notice that the coefficient of x^2 and y^2 and their signs are both different. That means that the given in the problem is equation of hyperbola.

Also, since it is the y^2 that is negative, then the equation must be converted to standard form of hyperbola with horizontal transverse axis.

The standard form is:

`(x-h)^2/a^2-(y-k)^2/b^2=1`

where (h,k) is the center, a represents the semi-transverse axis and b represents the semi-conjugate axis.

To transform it to standard equation, group the terms with x. Do the same for the terms with y.

`(3x^2+6x) - (12y^2-48y)-93=0`

Then, do the completing the square method for each group.

`3(x^2+2x)-12(y^2-4y)-93=0`

`3(x^2+2x+1)-12(y^2-4y+4)-93-3+48=0`

`3(x+1)^2-12(y-2)^2-48=0`

`3(x+1)^2-12(y-2)^2=48`

`(3(x+1)^2)/48-(12(y-2)^2)/48=48/48`

`(x+1)^2/16-(y-2)^2/4=1`

Hence, the center of the hyperbola is `(-1,2)` .

(b) Vertices

Note that the transverse axis of the hyperbola is horizontal. So, its vertices would have a coordinates of `(h+-a,k)` .

Since the values of h and k are known already, it is only the value of a that has to be determined.

To do so, consider the denominator of the first fraction of the standard equation which is 16. This means that:

`a^2=16`

`sqrt(a^2)=+-sqrt16`

`a=+-4`

Substituting values of h, k and a the formula of vertices of hyperbola, the coordinates would be:

(h+a,k)=(-1+4,2)=(3,2)

(h-a,k)=(-1-4,2)=(-5,2)

Thus, the vertices of the hyperbola are `(-5,2)` and `(3,2)` .

(c) Slopes of the Asymptotes.

The formula for the asymptotes of hyperbola with horizontal transverse axis is:

y=+-b/a(x-h)+k

So the slopes of the asymptotes are +-b/a.

Since the value of b is still not known, consider the denominator of the second fraction which is 4. This means that:

`b^2=4`

`sqrt(b^2)=+-sqrt4`

`b=+-2`

So plugging in the values of a, and b to the formula slope yields:

`m=+-b/a=+-2/4=+-1/2`

Hence, the slopes of the two asymptotes are `-1/2` and `1/2` .    

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