area of the region bounded by the hyperbola?

**shah4u19** · September 9th 2009, 02:10 PM

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.

**seld** · September 9th 2009, 02:47 PM

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.

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

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?

**shah4u19** · September 9th 2009, 02:55 PM

Thanks! But do i substitute x = 5 sec d(theta), If yes then when I continue, I get stuck on the way.

**seld** · September 9th 2009, 03:07 PM

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

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

Substituting when you take the integral

$x = 5sec(\theta)$

well what's $x^2$ ?

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

what if you let u = 2x?

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

when you substitute that in it gets messy.

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

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.

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