# Thread: Centeral conic question.

1. ## 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 $\pm4$ and that the asymptotes are $y=\frac{5x}{4}$and $-\frac{}{5x}{4}$
Graph
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.: . I am very lost as to what to do with this one though.
When I try to determine 'c' I come out with
I don't really see how that can be correct.

Thank you.

2. 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 $\pm4$ and that the asymptotes are $y=\frac{5x}{4}$and $-\frac{}{5x}{4}$
Graph
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.: . I am very lost as to what to do with this one though.
When I try to determine 'c' I come out with
I don't really see how that can be correct.

Thank you.
Hi quikwerk,

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

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

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

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

$y=\pm \frac{5}{4}x$

3. 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 $y=\pm\frac{5}{4}\sqrt{x^2-16}$and $x=\pm\frac{4}{5}\sqrt{y^2-25}$?

4. 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 $y=\pm\frac{5}{4}\sqrt{x^2-16}$and $x=\pm\frac{4}{5}\sqrt{y^2-25}$?
My guess would be the extents of the graph would be the Domain and Range.

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

Does that mean that $\mid x \mid\geq 4$ and $\mid y \mid\geq 5$?

6. Originally Posted by quikwerk
Does that mean that $\mid x \mid\geq 4$ and $\mid y \mid\geq 5$?
The Domain would be $\{x\;|\;x \le -4 \r\:x \ge 4\}" alt="\{x\;|\;x \le -4 \r\:x \ge 4\}" />

The Range would be $\{y\;|\; y \in \Re\}$

7. Two points: To find the "asymptotes", think about what happens for x and y really, really big! For $\frac{x^2}{16}$ and $\frac{y^2}{25}$ are far larger than the "1" on the right side- if we ignore, it we see that the curve must be close to $\frac{x^2}{16}- \frac{y^2}{25}= \left(\frac{x}{4}- \frac{y}{5}\right)\left(\frac{x}{4}+ \frac{y}{5}\right)= 0$ which means that either $\frac{x}{4}- \frac{y}{5}= 0$ so that $4y= 5x$ or $\frac{x}{4}+ \frac{y}{5}= 0$ so that $4y= -5x$. Those are the asymptotes.

For the "extent" try solving for x and y. $\frac{y^2}{25}= \frac{x^2}{16}- 1$ so that $y= \pm 5\sqrt{\frac{x^2}{16}-1}$. Since the argument of the square root must be positive, we must have $\frac{x^2}{16}- 1\ge 0$ or $x^2\ge 16$ so that $x\le -4$ or $x\ge 4$.
$x= \pm 4\sqrt{\frac{y^2}{25}+ 1}$. Since the argument is always positive, there is no restriction on y.

8. Thank you guys! I'll try to work it out with your advice and report back later.