# area of the region bounded by the hyperbola?

• Sep 9th 2009, 02:10 PM
shah4u19
area of the region bounded by the hyperbola?
Please help me with this problem. I have been trying 2 do it 4 half an hour and still can figure out. Thank You!

Find the area of the region bounded by the hyperbola 25x2 - 4y2 = 100 and the line x = 3.
• Sep 9th 2009, 02:47 PM
seld
So the area looks a little like a slice of a circle right?

the easiest way to do it is turn y in terms of x.

$\displaystyle 25x^2-4y^2=100$
$\displaystyle 25x^2-100-4y^2=0$
$\displaystyle 25x^2-100=-4y^2$
$\displaystyle \frac{25}{4}x^2-25=y^2$
$\displaystyle \sqrt{\frac{25}{4}x^2-25}=y$

Since you need both the positive and negative portion instead of doing 2 integrals you can just take the integral to get the region above the x axis and multiply it's area by 2.

can you take it from there?
• Sep 9th 2009, 02:55 PM
shah4u19
Thanks! But do i substitute x = 5 sec d(theta), If yes then when I continue, I get stuck on the way.
• Sep 9th 2009, 03:07 PM
seld
$\displaystyle \sqrt{\frac{25}{4}x^2-25}=y$

$\displaystyle \int\sqrt{\frac{25}{4}x^2-25}dx$

Substituting when you take the integral

$\displaystyle x = 5sec(\theta)$

well what's $\displaystyle x^2$?

$\displaystyle x^2=25sec^2(\theta)$ right?

what if you let u = 2x?

and $\displaystyle dx=5sec\theta tan\theta \text{ }d\theta$ right?

when you substitute that in it gets messy.

$\displaystyle \sqrt{\frac{25}{4}x^2-25}=y$
$\displaystyle =\sqrt{(25)*(\frac{1}{4}x^2-1)}$
$\displaystyle =5\sqrt{\frac{x^2}{4}-1}$

what if you let u = 2x?

Seems like you should do a trig substitution after that, though your initial idea to substitute secant isn't wrong, just needs somewhat different coefficients.