Given the equation 3x^2 - 12y^2 + 6x +48y - 93 = 0 determine: a)  The Center C b)  The 2 Vertices c)  The slopes of the asymptotes (enter as a reduced fraction)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The given equation 3x^2 - 12y^2 + 6x +48y - 93 = 0 can be written as:

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

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

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

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

This is the equation of a hyperbola with center (-1, 2). The vertices are at `(-1+-2, 2)`
The slopes of the asymptotes is `+-1/2`

The equation of the hyperbola is `(x + 1)^2/4^2 - (y - 2)^2/2^2 = 1` , center at `(-1, 2)` , vertices at `(-1+-2, 2)` and the slope of the asymptote `+-1/2`

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