# Thread: Two central conic questions.

1. ## Two central conic questions.

I have two questions for which I have partial working for, but I am unsure of the working for them.

Q)1:
Find (a) the directrix, (b) the focus and (c) the roots of the parabola $y=x^2-5x+4$

a)
y= $-\frac{5}{2}$

b)
(2.5,-2)

c)
$y=x^2-5x+4\rightarrow y=(x-4)(x-1)$
Therefore x=4 and x=1.

I'm very certain of my answer for c), however I lost my working for a) and b) so I'm fairly certain that I am incorrect for those answers.

Q)2
Graph $\frac{x^2}{16}-\frac{y^2}{25}=1$
Show how you arrived at your graph by determining the x-intercepts, the extent of the graph and the asympotes.

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.: $\frac{y^2}{25}-\frac{x^2}{16}$. I am very lost as to what to do with this one though.
When I try to determine 'c' I come out with $\sqrt{41}$
I don't really see how that can be correct.

Thank you.

2. Completing squares:

$y=\left(x-5/2\right)^2-25/4+4$

or

$y-(-9/4)=4(1/4)(x-5/2)^2$

we have the parabola in the canonical form:

$(y-\beta)=4f(x-\alpha)^2$

Then, the directrix is:

$y=\beta-f=-9/4-1/4=-5/2$

etc.

Fernando Revilla

3. I'm not at all expireienced with canonical form of parabolas...so I dont really understand what you are saying...

4. Originally Posted by quikwerk
I'm not at all expireienced with canonical form of parabolas...so I dont really understand what you are saying...

Well, that is a standard way. For helping you I should know what can I use.

Fernando Revilla

.

5. I can use completing the square and other high school algebra methods.

6. Originally Posted by quikwerk
I'm not at all expireienced with canonical form of parabolas...so I dont really understand what you are saying...
Hi quikwerk,

The standard (or vertex) form of your parabola is $y=a(x-h)^2+k$

where (h, k) is your vertex.

The directrex is $y=k-\frac{1}{4a}$

To put your equation into vertex form, you complete the square like FernandoRevilla showed you.

$y=x^2-5x+4$

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

$y=(x-\frac{5}{2})^2-\frac{9}{4}$

Your directrix is $y=-\frac{9}{4}-\frac{1}{4(1)}$

$y=-\frac{5}{2}$

7. Many thanks, both of you!
Masters, I finally understand how to work this question.

Can you help me with the second question?

I have determined the x-intercepts to be 4 and -4 and I have figured out the asymptotes as $y=\pm\frac{5x}{4}$

8. The focus in Q1 can be found using this little ditty:

$F\left(h\;,\;k+\frac{1}{4a}\right)$

9. Originally Posted by quikwerk
Many thanks, both of you!
Masters, I finally understand how to work this question.

Can you help me with the second question?

I have determined the x-intercepts to be 4 and -4 and I have figured out the asymptotes as $y=\pm\frac{5x}{4}$
Post Q2 as a separate post and someone will get to it. I promise. I'm going to be tied up for about an hour.

10. Thanks again Masters!