# Math Help - ellipses

1. ## ellipses

Sketch the graphs of the following function:

9x^2-7x+y^2-4y+112=0

Any help/or complete working out would be appreciated!
I have tried to do this and failed.

2. Use completing the square to change the function into something like:
9(x - a)^2 + (y - b)^2 = c

Can you first figure out where the center of the ellipse is?

3. Hello, Joker37!

I bet that the "112" should be negative . . . even then, the answer is ugly!

You were assigned this problem with absolutely no instructions?
How cruel!

$\text{Sketch the graph of the following function: }\;9x^2-7x+y^2-4y-112\:=\:0$

We have: . $9(x^2 - \frac{7}{9}x \quad) + (y^2 - 4y \quad) \;=\;112$

Complete the square: . $9\left(x^2 - \frac{7}{9}x + \frac{49}{324} - \frac{49}{384}\right) + (y^2 - 4y + 4 - 4) \;=\;112$

. . $9\left(x^2 - \frac{7}{9}x + \frac{49}{324}\right) - \frac{49}{36} + (y^2 - 4y + 4) - 4 \:=\:112$

. . . . . . $9\left(x - \frac{7}{18}\right)^2 + (y - 2)^2 \;=\;112 + \frac{49}{36} + 4$

. . . . . . $9\left(x-\frac{7}{18}\right)^2 + (y-2)^2 \;=\;\frac{4225}{36}$

Divide by $\frac{4225}{36}$

. . $\displaystyle \frac{9(x-\frac{7}{18})^2}{\frac{4225}{36}} + \frac{(y-2)^2}{\frac{4225}{36}} \;=\;1$

. . $\displaystyle \frac{(x - \frac{7}{18})^2}{\frac{4225}{324}} + \frac{(y-2)^2}{\frac{4225}{36}} \;=\;1$

. . $\displaystyle \frac{(x-\frac{7}{18})^2}{\left(\frac{65}{18}\right)^2} + \frac{(y-2)^2}{\left(\frac{65}{6}\right)^2} \;=\;1$

$\text{Center: }\:\left(\frac{7}{18},\:2\right)$

. . $a = \frac{65}{6} \;\text{ (vertical)}$

. . $b = \frac{65}{18}\;\text{ (horizontal)}$

4. Actually, as it is written, it is very easy- the graph is the empty set!

5. Originally Posted by Soroban
Hello, Joker37!

I bet that the "112" should be negative . . . even then, the answer is ugly!

You were assigned this problem with absolutely no instructions?
How cruel!

We have: . $9(x^2 - \frac{7}{9}x \quad) + (y^2 - 4y \quad) \;=\;112$

Complete the square: . $9\left(x^2 - \frac{7}{9}x + \frac{49}{324} - \frac{49}{384}\right) + (y^2 - 4y + 4 - 4) \;=\;112$

. . $9\left(x^2 - \frac{7}{9}x + \frac{49}{324}\right) - \frac{49}{36} + (y^2 - 4y + 4) - 4 \:=\:112$

. . . . . . $9\left(x - \frac{7}{18}\right)^2 + (y - 2)^2 \;=\;112 + \frac{49}{36} + 4$

. . . . . . $9\left(x-\frac{7}{18}\right)^2 + (y-2)^2 \;=\;\frac{4225}{36}$

Divide by $\frac{4225}{36}$

. . $\displaystyle \frac{9(x-\frac{7}{18})^2}{\frac{4225}{36}} + \frac{(y-2)^2}{\frac{4225}{36}} \;=\;1$

. . $\displaystyle \frac{(x - \frac{7}{18})^2}{\frac{4225}{324}} + \frac{(y-2)^2}{\frac{4225}{36}} \;=\;1$

. . $\displaystyle \frac{(x-\frac{7}{18})^2}{\left(\frac{65}{18}\right)^2} + \frac{(y-2)^2}{\left(\frac{65}{6}\right)^2} \;=\;1$

$\text{Center: }\:\left(\frac{7}{18},\:2\right)$

. . $a = \frac{65}{6} \;\text{ (vertical)}$

. . $b = \frac{65}{18}\;\text{ (horizontal)}$
Haha, well it says +112.
However the simplified answer for the equation is supposed to be 36(x-4)^2 + 4(y-2)^2 = 144.

Soroban, I really appreciate your help and efforts so far!

6. Hi everybody!
I'm still stuck with this problem and unfortunately I have no-body to help me. I feel so alone.

Anyway, if it's because the question doesn't make sense or something like that please tell me and I will skip this question but I just don't like knowing that I don't know how to do something.

7. Originally Posted by Joker37
Haha, well it says +112.
However the simplified answer for the equation is supposed to be 36(x-4)^2 + 4(y-2)^2 = 144.

Soroban, I really appreciate your help and efforts so far!

If you look carefully at the equations used, you will notice that $36(x-4)^2 + 4(y-2)^2 = 144$ and $9x^2-7x+y^2-4y+112=0$ are different equations.

If you expand and simplify the simplified one, you will get: $9 x^2-72 x+y^2-4 y+112 = 0$ which I'm guessing is the equation you are trying to graph and that the one in the first post has a typing error?

Now that Soroban has supplied you with a method, why don't you give it a try yourself?

8. Hello, Joker37!

$\text{Haha, well it says +112.}$

$\text{However the simplified answer for the equation is supposed to be:}$
. . $36(x-4)^2 + 4(y-2)^2 \:=\: 144$

Well, that certainly explains a lot!

First of all, the original equation had a typo
. . It was not 7x, but 72x.

Second, their answer is quite stupid.
. . That is definitely not the simplfied answer.

$\begin{array}{ccccccc}\text{Neither is:} &18(x-4)^2 + 2(y-2)^2 \:=\:72 \\
\text{nor:} & 360(x-4)^2 + 40(y-2)^2 \:=\:1440 \end{array}$