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)` .
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` .