Author Topic: Hyperbolas (Read 1269 times) Tweet Share

darkphoenix · « **on:** February 15, 2010, 11:56:36 pm »

Find the asymptotes for the hyperbola $\frac{2x^2}3 - \frac{y^2}2 = 1$

Not sure exactly how the 2 in front of the $x^2$ changes it.

kyzoo · « **Reply #1 on:** February 16, 2010, 12:07:16 am »

$\frac{2x^2}{3}<br />=\frac{x^2}{\frac{3}{2}}<br />=\left ( \sqrt{\tfrac{x}{\left ( \tfrac{3}{2} \right )}} \right )^2<br />$

Therefore it is dilated parallel to the x-axis by factor $\sqrt{\frac{3}{2}}$

QuantumJG · « **Reply #2 on:** February 16, 2010, 12:30:10 am »

Quote from: darkphoenix on February 15, 2010, 11:56:36 pm

Find the asymptotes for the hyperbola $\frac{2x^2}3 - \frac{y^2}2 = 1$

Not sure exactly how the 2 in front of the $x^2$ changes it.

I did a detailed thing on hyperbolas here (http://vcenotes.com/forum/index.php/topic,23055.msg234099.html#msg234099).

Anyway,

The asyptotes are

$y = \pm \dfrac{ 2 \sqrt{3} }{3} x$

darkphoenix · « **Reply #3 on:** February 16, 2010, 12:37:56 am »

Quote from: QuantumJG on February 16, 2010, 12:30:10 am

Quote from: darkphoenix on February 15, 2010, 11:56:36 pm
Find the asymptotes for the hyperbola $\frac{2x^2}3 - \frac{y^2}2 = 1$

Not sure exactly how the 2 in front of the $x^2$ changes it.

I did a detailed thing on hyperbolas here (http://vcenotes.com/forum/index.php/topic,23055.msg234099.html#msg234099).

Anyway,

The asyptotes are

$y = \pm \dfrac{ 2 \sqrt{3} }{3} x$

Yeah i got that too, but my book says:

$y = \pm \dfrac{4}{\sqrt3} x$

QuantumJG · « **Reply #4 on:** February 16, 2010, 12:57:36 am »

Quote from: darkphoenix on February 16, 2010, 12:37:56 am

Quote from: QuantumJG on February 16, 2010, 12:30:10 am
Quote from: darkphoenix on February 15, 2010, 11:56:36 pm
Find the asymptotes for the hyperbola $\frac{2x^2}3 - \frac{y^2}2 = 1$

Not sure exactly how the 2 in front of the $x^2$ changes it.

I did a detailed thing on hyperbolas here (http://vcenotes.com/forum/index.php/topic,23055.msg234099.html#msg234099).

Anyway,

The asyptotes are

$y = \pm \dfrac{ 2 \sqrt{3} }{3} x$

Yeah i got that too, but my book says:

$y = \pm \dfrac{4}{\sqrt3} x$

Sure they didn't typo?

$y = \pm \dfrac{b}{a} x$

$b = \sqrt{2}$ & $a = \dfrac{ \sqrt{3} }{ \sqrt{2} }$

$\therefore y = \pm \dfrac{ \sqrt{2} }{1} \times \dfrac{ \sqrt{2} }{ \sqrt{3} } x = \pm \dfrac{ \sqrt{4} }{ \sqrt{3} } x$

darkphoenix · « **Reply #5 on:** February 16, 2010, 12:23:49 pm »

Hm not sure if its typo.

Thats just what the answers in the back of the book say..

superflya · « **Reply #6 on:** February 16, 2010, 05:16:32 pm »

typo.. i got $y=\pm \frac{2 \sqrt{3}}{ 3}$

darkphoenix · « **Reply #7 on:** February 19, 2010, 03:41:57 pm »

kk cool

Stroodle · « **Reply #8 on:** February 19, 2010, 04:19:34 pm »

Sux how there's quite a few errors in the new Maths Quest text...

darkphoenix · « **Reply #9 on:** February 19, 2010, 05:34:07 pm »

Yeah, which further proves how shit it is.

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Author Topic: Hyperbolas (Read 1269 times) Tweet Share

darkphoenix

Hyperbolas

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