# Centeral conic question.

• Jan 27th 2011, 09:09 AM
quikwerk
Centeral conic question.
I'm not too sure of how to work this one out, I'm really having trouble determining the extent of the graph. I have figured out that the x intercept is $\displaystyle \pm4$ and that the asymptotes are $\displaystyle y=\frac{5x}{4}$and$\displaystyle -\frac{}{5x}{4}$
Graph http://www.mathhelpforum.com/math-he...ba2ea7f964.png
Show how you arrived at your graph by determining the x-intercepts, the extent of the graph and the asymptotes.

A) I admit to being lost when it comes to this question, I know how to work these kind of questions under other situations where the x and y are reversed. E.g.: http://www.mathhelpforum.com/math-he...14fcf83633.png. I am very lost as to what to do with this one though.
When I try to determine 'c' I come out with http://www.mathhelpforum.com/math-he...9a8ccf57fa.png
I don't really see how that can be correct.

Thank you.
• Jan 27th 2011, 09:35 AM
masters
Quote:

Originally Posted by quikwerk
I'm not too sure of how to work this one out, I'm really having trouble determining the extent of the graph. I have figured out that the x intercept is $\displaystyle \pm4$ and that the asymptotes are $\displaystyle y=\frac{5x}{4}$and$\displaystyle -\frac{}{5x}{4}$
Graph http://www.mathhelpforum.com/math-he...ba2ea7f964.png
Show how you arrived at your graph by determining the x-intercepts, the extent of the graph and the asymptotes.

A) I admit to being lost when it comes to this question, I know how to work these kind of questions under other situations where the x and y are reversed. E.g.: http://www.mathhelpforum.com/math-he...14fcf83633.png. I am very lost as to what to do with this one though.
When I try to determine 'c' I come out with http://www.mathhelpforum.com/math-he...9a8ccf57fa.png
I don't really see how that can be correct.

Thank you.

Hi quikwerk,

You are correct. $\displaystyle c=\sqrt{41}$

The standard form for this hyperbola is $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The equations of the asymptotes are $\displaystyle y-k=\pm \frac{b}{a}(x-h)$

$\displaystyle y-0=\pm \frac{5}{4}(x-0)$

$\displaystyle y=\pm \frac{5}{4}x$
• Jan 27th 2011, 10:06 AM
quikwerk
Thanks- still not sure how to work out the extent of the graph : (

Am i correct in thinking that the extent of the graph is given by the equations$\displaystyle y=\pm\frac{5}{4}\sqrt{x^2-16}$and$\displaystyle x=\pm\frac{4}{5}\sqrt{y^2-25}$?
• Jan 27th 2011, 10:31 AM
masters
Quote:

Originally Posted by quikwerk
Thanks- still not sure how to work out the extent of the graph : (

Am i correct in thinking that the extent of the graph is given by the equations$\displaystyle y=\pm\frac{5}{4}\sqrt{x^2-16}$and$\displaystyle x=\pm\frac{4}{5}\sqrt{y^2-25}$?

My guess would be the extents of the graph would be the Domain and Range.
• Jan 27th 2011, 10:44 AM
quikwerk
Quote:

Originally Posted by masters
My guess would be the extents of the graph would be the Domain and Range.

Does that mean that $\displaystyle \mid x \mid\geq 4$ and $\displaystyle \mid y \mid\geq 5$?
• Jan 27th 2011, 11:15 AM
masters
Quote:

Originally Posted by quikwerk
Does that mean that $\displaystyle \mid x \mid\geq 4$ and $\displaystyle \mid y \mid\geq 5$?

The Domain would be $\displaystyle \{x\;|\;x \le -4 \:or\:x \ge 4\}$

The Range would be $\displaystyle \{y\;|\; y \in \Re\}$
• Jan 27th 2011, 02:53 PM
HallsofIvy
Two points: To find the "asymptotes", think about what happens for x and y really, really big! For $\displaystyle \frac{x^2}{16}$ and $\displaystyle \frac{y^2}{25}$ are far larger than the "1" on the right side- if we ignore, it we see that the curve must be close to $\displaystyle \frac{x^2}{16}- \frac{y^2}{25}= \left(\frac{x}{4}- \frac{y}{5}\right)\left(\frac{x}{4}+ \frac{y}{5}\right)= 0$ which means that either $\displaystyle \frac{x}{4}- \frac{y}{5}= 0$ so that $\displaystyle 4y= 5x$ or $\displaystyle \frac{x}{4}+ \frac{y}{5}= 0$ so that $\displaystyle 4y= -5x$. Those are the asymptotes.

For the "extent" try solving for x and y. $\displaystyle \frac{y^2}{25}= \frac{x^2}{16}- 1$ so that $\displaystyle y= \pm 5\sqrt{\frac{x^2}{16}-1}$. Since the argument of the square root must be positive, we must have $\displaystyle \frac{x^2}{16}- 1\ge 0$ or $\displaystyle x^2\ge 16$ so that $\displaystyle x\le -4$ or $\displaystyle x\ge 4$.
$\displaystyle x= \pm 4\sqrt{\frac{y^2}{25}+ 1}$. Since the argument is always positive, there is no restriction on y.
• Jan 27th 2011, 03:51 PM
quikwerk
Thank you guys! I'll try to work it out with your advice and report back later.