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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lemjay | High School Teacher | (Level 2) Senior Educator

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(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:


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.







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:




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



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:


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:




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


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

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