Sketch polar curve (conic i believe)

[jwxie]

Homework Statement

Let C be the curve whose equation is given by: 4x^2+18y-20=9y^2+16x+9

Identify and sketch the curve C and if possible give its center, directrix, foci, vertices, and endpoints on the minor axis.

Homework Equations

I have learned parabola, hyperbola and elispes (these are conic sections)

The Attempt at a Solution

To start with, I really can't. I tried to rearrange the equations.
4x^2 - 9y^2+18y-16x=29
but i don't see how i can graph this... please guide me through this, thank you.

edited
i just asked wolfram to do graph 4x^2 - 9y^2+18y-16x=29 and i got two hyperbolas (they are not symmetrics)... but how come? i don't see the 1 in any side

 

[Cyosis]

Try to get rid of the first order x and y terms by completing the square. This way you can cast the equation into the form $a(x-b)^2-c(y-d)^2=e$, which should be familiar.

 

[jwxie]

so 4x^2+18y-20=9y^2+16x+9
i get 4(x^2 - 4x + 4) - 9(x^2 + 2y +1) = 29
so
4(x-2)^2 - 9(x+1)^2 = 29

is that correct?
this is a parabola.

but this 4(x-2)^2 - 9(x+1)^2 = 29 produces a different graph than the original form in wolfram.
i did get a parabola, but using the original form i got something looked like hyperbola

i got really confused because how can i tell what the equation supposes to be?
like in the case 9x^2+y^2=35, the ans key said it was an ellipse...

 

[Mark44]

so 4x^2+18y-20=9y^2+16x+9
i get 4(x^2 - 4x + 4) - 9(x^2 + 2y +1) = 29
so
4(x-2)^2 - 9(x+1)^2 = 29

is that correct?

No, for several reasons.
Your original equation involved both x and y. Your new equation does not. Also, when you completed the square, it looks like you forgot to add the same quantities to the right side. For example, when you added 4 inside the parentheses to complete the square in x^2 -4x, you are really adding 16. You need to add 16 to the right side also.

this is a parabola.

but this 4(x-2)^2 - 9(x+1)^2 = 29 produces a different graph than the original form in wolfram.
i did get a parabola, but using the original form i got something looked like hyperbola

That's because it is a hyperbola.

i got really confused because how can i tell what the equation supposes to be?
like in the case 9x^2+y^2=35, the ans key said it was an ellipse...

 

[Cyosis]

so 4x^2+18y-20=9y^2+16x+9
i get 4(x^2 - 4x + 4) - 9(x^2 + 2y +1) = 29
so
4(x-2)^2 - 9(x+1)^2 = 29is that correct?
this is a parabola.

You're interchanging x-es and y-s at random. That is why you get a parabola in your final expression. If you do it correctly you will get a hyperbola. You need to arrive at an equation of the form in post #2 which includes y.

i got really confused because how can i tell what the equation supposes to be?
like in the case 9x^2+y^2=35, the ans key said it was an ellipse...

You should learn the 'definitions' of a hyperbola and an ellipse. I am pretty sure you will find them in your textbook.

Ellipse
$$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$$

Hyperbola
$$\frac{(x-x_0)^2}{a^2}-\frac{(y-y_0)^2}{b^2}=1$$

Whenever you have a problem like this you check if you can cast your equation into one of the two forms above.

 

[jwxie]

oh my bad. it was a typo, no wonder why
4(x-2)^2 - 9(y+1)^2 = 29
now i have to get it back to the form
with a^2 and b^2 as fraction and 1 as the difference
how can i do that?
our textbook only have examples with the regular form -_-
do i just divide everything by 29? and so a = sqrt(29)/2 and b = sqrt(29)/3

=_=
(((x-2)^2)/(sqrt(29)/2)^2) - (((y+1)^2)/(sqrt(29)/3)^2) = 1

it seems right to me in wolfram... what a complicated form...

 

[Cyosis]

Yes that is correct.

 

[Mark44]

I beg to differ. It is not correct.

4x² + 18y - 20 = 9y² + 16x + 9
<==> 4x² - 16x - 9y² + 18y = 29
<==> 4(x² - 4x + ?) - 9(y² - 2y + ?) = 29

Can you continue from here? There are at least two errors in your work - forgetting to change the sign of 18y when you factor out -9, and forgetting to add the right amounts to the right side after completing the square.

When you're done, the number on the right side should NOT be 29.

 

[jwxie]

omg. i am so dead right now. i kept putting the wrong number. on the draft paper i did get -1 for the y

oh i missed your second comment

 

[Mark44]

No, the 29 is wrong. There is no 29 in the final equation. In the last line of my previous post I had
<==> 4(x² - 4x + ?) - 9(y² - 2y + ?) = 29

To complete the square in the x terms, add 4. This means I am really adding 16 to the left side because of the multiplier 4 in front of the group, so I have to add 16 to the right side of the equation. When you complete the square in the y terms, you have to do something similar.

If I were you I would go back and review completing the square.

 

[jwxie]

4(x^2 - 4x + 4) - 16 - 9(y2 - 2y + 1) - 9= 29 + 16 - 9
am i getting 36?

 

[Mark44]

4(x^2 - 4x + 4) - 16 - 9(y2 - 2y + 1) - 9= 29 + 16 - 9
am i getting 36?

This is still wrong. THINK ABOUT WHAT YOU ARE DOING!

On the left side you are adding 16 and then subtracting 16. You are also subtracting 9 and then subtracting 9 again.

 

[jwxie]

This is still wrong. THINK ABOUT WHAT YOU ARE DOING!

On the left side you are adding 16 and then subtracting 16. You are also subtracting 9 and then subtracting 9 again.

i thought the - carries to all the 9's there

4(x^2 - 4x + 4) 16 - 9(y2 - 2y + 1) - 9= 29 + 16 + 9
i really got confused here.

when i did it in wolfram, complete square (the original form)
i got this
4 (x-2)^2-9 (y-1)^2-36 = 0

 

[Char. Limit]

Why would you add nine to the right side? The 9 on the left side is subtracted, not added...

 

[Mark44]

i thought the - carries to all the 9's there

4(x^2 - 4x + 4) 16 - 9(y2 - 2y + 1) - 9= 29 + 16 + 9
i really got confused here.

Well, yes, but why are you confused? This isn't complicated.

4(x² - 4x + ?) - 9(y² - 2y + ?) = 29
<==> 4(x² - 4x + 4) - 9(y² - 2y + ?) = 29 + 16
I added 16 (= 4 * 4) to the left side, so to keep the equation equivalent to the one before it I added 16 to the right side. Can you work the rest of it by yourself (without Wolfram Alpha)?

when i did it in wolfram, complete square (the original form)
i got this
4 (x-2)^2-9 (y-1)^2-36 = 0

Until you understand completing the square, I would advise you not to use WA for this purpose. Besides, you are not likely to be able to use it for a test.

 

[jwxie]

Hi Mark
I believe what you are saying is that whatever i added, i do the same on the other side
whatever i substracted, i do the same on the other side
since i added 16, i added 16 to 29
i minused 9 so i -9

am i correct there?

 

[Mark44]

Yes, that's exactly what I'm saying. Continuing the process from here -
4(x² - 4x + 4) - 9(y² - 2y + ?) = 29 + 16

what do you get?

 

[jwxie]

4(x2 - 4x + 4) - 9(y2 - 2y + 1) (-9) = 29 + 16 -9
36?

 

[Mark44]

4(x2 - 4x + 4) - 9(y2 - 2y + 1) (-9) = 29 + 16 -9
36?

Why is this.........^... here? I.e., (-9).

 

[jwxie]

i thought you agreed that
4(x2 - 4x + 4) - 9(y2 - 2y + 1) (+1 * -9) = ...
I am sorry if I still not able to answer the question .correctly ...

 

[Mark44]

No, I never agreed to any such thing. If you think I did, please say which post you think I said this. I'm not even sure what you mean by it. Are you multiplying -9(y² - 2y + 1) by -9? That makes no sense at all.

 

[Mark44]

Here's a similar example, step by step. Normally I would multiply both sides by -1, but for the purposes of this example I'll leave things as they are.

-3y² - 12y = -15
<==> -3(y² + 4y) = -15
<==> -3(y² + 4y + 4) = -15 -12 --- Notice here that I added -12 to both sides
<==> -3(y + 2)² = -27

For your problem, you'll stop at an equation similar to the above, but I'll continue my example the rest of the way.

<==> (y + 2)² = 9
<==> y + 2 = 3 or y + 2 = -3
<==> y = 1 or y = -5