# Find the centre, foci and asymptotes of the hyperbola [closed]

Hyperbola:

$4x^2 - 8x - y^2 + 6y - 1 = 0$

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## closed as off-topic by N. F. Taussig, Davide Giraudo, abiessu, drhab, Hagen von EitzenJul 30 at 17:42

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Is this a homework question? –  Ron Gordon Jan 6 '13 at 21:32
It's a practice question from an exercise sheet. –  neverloggedin Jan 6 '13 at 21:36

For asymptotes: First of all try to form the equation as: $$\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1$$ for some suitable value $a,h,b,k$. Then the asymptotes of hyperbola are as $$y - k = \pm \dfrac{b}{a}(x - h)$$

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Nice w o r k +1 –  amWhy Feb 22 '13 at 2:06

See Hyperbola for help in identifying the center and foci of your hyperbola. There is also more information for the various formulae for a hyperbola, depending on where it is centered, and/or an East-West Opening hyperbola, or a North-South opening parabola.

We work to get the equation into the "standard form" of a hyperbola: $$\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1\tag{1}$$ for some apppropriate values $a,h,b,k$.

To do this for your equation, complete the squares:

$$4x^2 - 8x + 4 - y^2 + 6y - 1 - 8 = 0 + 4 - 8 = -4 \tag{2}$$

$$4(x^2-2x+1) - (y^2-6y+9) =-4\tag{3}$$

$$4(x - 1)^2 - (y - 3)^2 = -4\tag{4}$$

$$\frac{(x - 1)^2}{1} - \frac{(y-3)^2}{4} = \frac{-4}{4} = -1\tag{5}$$

$$\frac{(x - 1)^2}{1^2} - \frac{(y-3)^2}{2^2} = -1\quad \iff \quad \frac{(y-3)^2}{2^2} - \frac{(x - 1)^2}{1^2} = 1\tag{6*}$$

Then the asymptotes of hyperbola are as $$y - k = \pm \dfrac{b}{a}(x - h)$$

Hint: From $(6^*)$, given the form of the equation on the right, we see that your equation is that of a hyperbola with its transverse axis aligned with the y-axis: i.e., it's a "North-South opening hyperbola". Graphing ALWAYS helps, if possible.

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Your answer made my short hint very clear, step by step. + :) –  Babak S. Jan 7 '13 at 5:48