Hyperbolas?

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May 9th 2011, 01:14 PM
homeylova223

Hyperbolas?

Write an equation of the hyperbola that meet each set of conditions

37. A vertex is at (4,5) the center at (4,2) and an equation of one asymptote is 4y+4=3x

I managed to get this far into writing the equation (y-2)^2/9+(x-4)^2/ but I am not sure how to get "b" in the equation.

2. The hyperbola as foci at (0,8) and (0,-8) and eccentricity is 4/3
I am not sure how to do this one(Surprised)
May 9th 2011, 01:24 PM
TheChaz

Quote:

Originally Posted by homeylova223

Write an equation of the hyperbola that meet each set of conditions

37. A vertex is at (4,5) the center at (4,2) and an equation of one asymptote is 4y+4=3x

I managed to get this far into writing the equation (y-2)^2/9+(x-4)^2/ but I am not sure how to get "b" in the equation.

2. The hyperbola as foci at (0,8) and (0,-8) and eccentricity is 4/3
I am not sure how to do this one(Surprised)

4y + 4 = 3x -->
4y - 8 = 3x - 12 .....(we need the left to be a multiple of y - 2, and the right a multiple of x - 4 )
4(y - 2) = 3(x - 4)
y - 2 = 3/4(x - 4)
Now 3/4 is either a/b or b/a, depending on whether this is vertical or horizontal hyperbola (I can't remember which at the moment, so I'll leave it at that).
May 9th 2011, 01:42 PM
homeylova223

Hmm I did not know about the multiple thing.
May 9th 2011, 01:50 PM
TheChaz

Quote:

Originally Posted by homeylova223

Hmm I did not know about the multiple thing.

Well in the formula for a hyperbola, you have - among other things - ... (x - h) and (y - k).
LIKEWISE, in the asymptote equations, you have (y - k) = m(x - h), where h and k are exactly as in the hyperbola equation. So they are multiples.
May 9th 2011, 04:43 PM
HallsofIvy

Since the center has both x components the same the equation is of the form $-\frac{(x- x_0)^2}{a^2}+ \frac{(y- y_0)^2}{b^2}= 1$ . With center at (4, 2), that is $-\frac{(x- 4)^2}{a^2}+ \frac{(y- 2)^2}{b^2}= 1$ . Such a hyperbola has asympotes satisfying $-\frac{(x- 4)^2}{a^2}+ \frac{(y- 2)^2}{b^2}= 0$ which is the same as $\frac{x- 4}{a^2}= \pm\frac{y- 2}{b}$ or $x= 4\pm\frac{a}{b}(y- 2)$ .

You are told that one asymptote is of the form 4y+ 4= 3x or $x= \frac{4}{3}+ \frac{4}{3}y= \frac{4}{3}+ \frac{4}{3}(y- 2)+ \frac{8}{3}= 4+ \frac{4}{3}(y- 2)$ .

Can you see what a and b must be now?
May 9th 2011, 05:18 PM
homeylova223

Yes I see the only trouble I am having right now is with the second question.
May 10th 2011, 10:20 AM
HallsofIvy

"The hyperbola as foci at (0,8) and (0,-8) and eccentricity is 4/3"
The center of the hyperbola is the midpoint if the segment between the two foci so the center of this hyperbola is at (0, 0) and it can be written as
$-\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$

Now, if f is the distance from (0, 0) to the foci (here 8), then $f^2= a^2+ b^2$ . The eccentricity, here 4/3, is given by $\sqrt{1+ \frac{a^2}{b^2}}= \frac{\sqrt{a^2+ b^2}}{b}$ .

That gives two equations to solve for a and b.

(I'll bet those formulas are in your textbook.)