1. ## hyperbole

can any one answer this,,,,really annoying me??? Im told to find the centre foci and hyperbola of $\displaystyle 4x^2-8x-y^2+6y-1=0$
I get everything except the asymptotes,, I think it involves limits or something because I dont get the conventional ax/b or was it bx/a because its switched around? Any way the foci was(1,3$\displaystyle \pm\sqrt{5})$ and centre (1,3) thanks!

2. Originally Posted by oxrigby
can any one answer this,,,,really annoying me??? Im told to find the centre foci and hyperbola of $\displaystyle 4x^2-8x-y^2+6y-1=0$
I get everything except the asymptotes,, I think it involves limits or something because I dont get the conventional ax/b or was it bx/a because its switched around? Any way the foci was(1,3$\displaystyle \pm\sqrt{5})$ and centre (1,3) thanks!
Since you know that the centre is (1,3), you can write the equation of the hyperbola as $\displaystyle 4(x-1)^2 - (y-3)^2 = - 4$. Now factorise the left side (difference of two squares): $\displaystyle \bigl(2(x-1) + (y-3)\bigr)\bigl(2(x-1) - (y-3)\bigr) = - 4$. If one of those factors on the left is very large, then the other one must be very small. That tells you that the asymptotes are the lines $\displaystyle 2(x-1) \pm(y-3) = 0$.

3. I presume you completed the square in $\displaystyle 4x^2-8x-y^2+6y-1=0$ to get $\displaystyle 4(x^2- 2x+ 4)- (y^2- 6y+ 9)= 16- 9+ 1= 18$ or $\displaystyle 4(x-2)^2- (y- 3)^2= 18$
In standard form, that is $\displaystyle \frac{(x- 2)^2}{\frac{9}{2}}- \frac{(y-3)^2}{18}= 1$.

Now, here's a technique that is worth knowing: For very large values of x and y, the "1" on the right will be negligible compared with the large numbers we are subtracting on the left. The graph will be very close to (technically, "asymptotic to") $\displaystyle \frac{(x- 2)^2}{\frac{9}{2}}- \frac{(y-3)^2}{18}= 0$.

That is equivalent to $\displaystyle \frac{(x-2)^2}{\frac{9}{2}}= \frac{(y-3)^2}{18}$. Taking square roots of both sides we get the two equations
$\displaystyle \frac{x-2}{\frac{3\sqrt{2}}{2}}= \frac{y-3}{3\sqrt{2}}$ and $\displaystyle \frac{x-2}{\frac{3\sqrt{2}}{2}}= -\frac{y-3}{3\sqrt{2}}$.

Those are the equations of the asymptotes to the hyperbola.