Classify the graph of a function

[krazydog]

I have an equation that i think is an ellipse or hyperbola since looking at the equation it has a \(\displaystyle y^ and a x^2 \).
However, i am having problems finishing it since i think i have to do complete the square but cant seem to get it.

Heres the equation.

\(\displaystyle 4x^2 + 4y^2 - 4x - 3 = 0 \)

Wouldn't i arrange like terms

\(\displaystyle 4x^2 -4x + 4y^2 - 3 = 0 \)

then, completing the square has me baffled.
First term i take half of \(\displaystyle -4x = 2^2 = 4\)

so i now have
\(\displaystyle (4x^2 -4x + 4) + (4y^2 - 3) = 0 + 4 \)

im not sure how to handle the \(\displaystyle 4y^2 - 3 \)

Wouldnt i get a fraction like \(\displaystyle 1/4 \) ?

From there i can divide out the 4 and -4 to get the implicit equation that defines an eclipse?

fillng in the right values from my equation.


\(\displaystyle \frac {(x - h)^2} { -4} + \frac {(y -k)^2}{4} = 1 \)

Thank you

 

[lookagain]

I have an equation that i think is an ellipse.
However, i am having problems finishing it since i think i have to do complete the square but cant seem to get it.

Heres the equation.

\(\displaystyle 4x^2 + 4y^2 - 4x - 3 = 0 \)

Wouldn't i arrange like terms

\(\displaystyle 4x^2 -4x + 4y^2 - 3 = 0 \)

then, completing the square has me baffled.
First term i take half of \(\displaystyle -4x = 2^2 = 4\)

You can't unless the coefficient of x^2 is 1.

\(\displaystyle Suggestion:\)

After your own rewrite of the original equation,
divide all terms on both sides of the equation by 4:


\(\displaystyle x^2 - x + y^2 - \dfrac{3}{4} = 0\)


\(\displaystyle Rewrite:\)


\(\displaystyle x^2 - x + y^2 = \dfrac{3}{4}\)


Take half of the coefficient on the x-term, and then square that result:


\(\displaystyle \bigg(\dfrac{-1}{2}\bigg)^2 = \ \dfrac{1}{4}\)


Add 1/4 to each side of the rewritten equation:


\(\displaystyle x^2 - x + \dfrac{1}{4} = 1\)


The expression on the left-hand side of the equation is a perfect square trinomial.

Factor that trinomial and insert it:


\(\displaystyle \bigg(x - \dfrac{1}{2}\bigg)^2 + y^2 = 1\)



\(\displaystyle This \ is \ in \ the \ form \ of \ \ (x - h)^2 + (y - k)^2 = r^2.\)



\(\displaystyle \text{What conic section is that?}\)

 

[krazydog]

wow, i forgot about dividing out! Anyways, i could follow along and it looks like a parabola?
not sure why you are using the r since polar coordinates are not covered with this problem..

Also, wouldnt i transfer the\(\displaystyle y^2\)
to the other side of the equation so they equal eachother?

So, from here i can find the vertex, focus, directrix!

Thank you

 

[wjm11]

wow, i forgot about dividing out! Anyways, i could follow along and it looks like a parabola?
not sure why you are using the r since polar coordinates are not covered with this problem..

Krazydog,

No, this is not a parabola, nor does the r have anything to do with polar coordinates in this problem.

Suggest you visit here for some great help:

http://www.purplemath.com/modules/conics.htm

"
...so also each conic has a "typical" equation form, sometimes along the lines of the following:

• parabola: Ax² + Dx + Ey = 0
  circle: x² + y² + Dx + Ey + F = 0
  ellipse: Ax² + Cy² + Dx + Ey + F = 0
  hyperbola: Ax² – Cy² + Dx + Ey + F = 0

These equations can be rearranged in various ways, and each conic has its own special form that you'll need to learn to recognize.."

Also, http://www.purplemath.com/modules/circle.htm

"
The "center-radius" form of the equation is:

• (x – h)² + (y – k)² = r²

...where the h and the k come from the center point (h, k) and the r² comes from the radius value r."

 

[krazydog]

thank you so much! The conic section didnt really cover the the circle in the homework so i guess that was further from my mind. I will defidently check out those links .. thank you very much!

 

[krazydog]

thank you