Hyperbola

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Sep 18th 2008, 12:37 PM
courteous

Hyperbola

Hyperbola is at the center of coordinate system. Its foci (?) are on abscissa. There are two points on the hyperbola; these are $T_1(\frac{3\sqrt5}{2},2)$ and $T_2(4,\frac{4\sqrt7}{9})$ . What is hyperbola's equation?
(I'm having trouble with solving two equations that follow(Headbang).)
Sep 18th 2008, 12:56 PM
Chris L T521

Quote:

Originally Posted by courteous

Hyperbola is at the center of coordinate system. Its foci (?) are on abscissa. There are two points on the hyperbola; these are $T_1(\frac{3\sqrt5}{2},2)$ and $T_2(4,\frac{4\sqrt7}{9})$ . What is hyperbola's equation?
(I'm having trouble with solving two equations that follow(Headbang).)

Since its foci are on the abscissa [x-axis] and the hyperbola is centered at the origin, we know that the hyperbola has the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , where $a>b$

Now plug the two points into this equation, and you will generate two new equations:

At $\left(4,\frac{4\sqrt{7}}{9}\right)$ , we get $\frac{16}{a^2}-\frac{112}{81b^2}=1$

At $\left(\frac{3\sqrt{5}}{2},2\right)$ , we get $\frac{45}{4a^2}-\frac{4}{b^2}=1$

After solving the system of equations, I got $a=\sqrt{\frac{981}{53}}$

But I'm getting $b^2=-\frac{1744}{171}\implies b\text{ is complex.}$

Are you sure that you wrote down the coordinates correctly?

--Chris
Sep 18th 2008, 10:19 PM
courteous

Quote:

Originally Posted by Chris L T521

Are you sure that you wrote down the coordinates correctly?

Sorry!!!(Doh) No!!! Sorry, Chris!(Worried) The points are $T_1(\frac{3\sqrt5}{2},2)$ and $T_2(4,\frac{4\sqrt7}{3})$ . The $T_2$ y-coordinate's denominator is 3 (not 9).
I've done it so many times that I've automatically restarted with partially already calculated number.(Speechless)