Graphing Conics

May 2010
10
0
4x^2 + 4y^2 +20x -16y + 37 =0
I completed the square and got:
(2x+10)^2 + (2y-8)^2 =127

x int ?
y int ?
Vertices ?
Center (-5, 4)
Domain ?
Range ?

I tried making y=0 to find the x intercepts but didn't work out. Same thing for y intercepts
Completely lost....
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, Honorable24!

\(\displaystyle 4x^2 + 4y^2 +20x -16y + 37 \:=\:0\)

I completed the square and got: .\(\displaystyle (2x+10)^2 + (2y-8)^2 \:=\:127\) .??

Find: .\(\displaystyle x\text{-intercepts,}\quad y\text{-intercepts,}\quad \text{Vertices,}\quad \text{Center,}\quad \text{Domain,}\quad \text{Range}\)

We have: .\(\displaystyle 4x^2 + 20x + 4y^2 - 16y \;=\;-37\)

. . . . \(\displaystyle 4(x^2 + 5x \quad) + 4(y^2 - 4y \quad) \;=\;-37\)

\(\displaystyle 4\left(x^2 + 5x + {\color{blue}\tfrac{25}{4}}\right)^2 + 4\left(y^2 - 4y + {\color{red}4}\right) \;=\;-37 + {\color{blue}25} + {\color{red}16} \)

. . . . . . . .\(\displaystyle 4\left(x + \tfrac{5}{2}\right)^2 + 4(y-2)^2 \;=\;4\)

. . . . . . . . . \(\displaystyle \left(x + \tfrac{5}{2}\right)^2 + (y-2)^2 \;=\;1\)


This is a circle: .\(\displaystyle \text{Center: }\left(-\tfrac{5}{2},\:2\right),\;\text{ radius } 2\)


There are no \(\displaystyle x\)-intercepts.

There are no \(\displaystyle y\)-intercepts.

A circle does not have vertices.

\(\displaystyle \text{Center: }\:\left(-\tfrac{5}{2},\:2\right)\)

\(\displaystyle \text{Domain: }\;x \in \left[-\tfrac{7}{2},\:-\tfrac{3}{2}\right] \)

\(\displaystyle \text{Range: }\;y \in[1,\:3] \)

 
May 2010
10
0
Thanks :)

x^2-6x + y^2 +4x=3

How do you do completing the square for this
I don't know how to do these types becuase there is no single y term
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Thanks :)

x^2-6x + y^2 +4x=3

How do you do completing the square for this
I don't know how to do these types becuase there is no single y term
I recommend you go back and check the problem. I suspect it should be \(\displaystyle x^2- 6x+ y^2+ 4y= 3\).

If \(\displaystyle x^2- 6x+ y^2+ 4x= 3\) really is the problem, -6x+ 4x= -2x so this is the same as \(\displaystyle x^2- 2x+ y^2= 3\). The fact that there is no "y to the first power" term is not a problem. It only means that y is already a "perfect square""\(\displaystyle y^2= (y- 0)^2\). Complete the square in x.
 
  • Like
Reactions: earboth

earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
Hello, Honorable24!


We have: .\(\displaystyle 4x^2 + 20x + 4y^2 - 16y \;=\;-37\)

. . . . \(\displaystyle 4(x^2 + 5x \quad) + 4(y^2 - 4y \quad) \;=\;-37\)

\(\displaystyle 4\left(x^2 + 5x + {\color{blue}\tfrac{25}{4}}\right)^2 + 4\left(y^2 - 4y + {\color{red}4}\right) \;=\;-37 + {\color{blue}25} + {\color{red}16} \)

. . . . . . . .\(\displaystyle 4\left(x + \tfrac{5}{2}\right)^2 + 4(y-2)^2 \;=\;4\)

. . . . . . . . . \(\displaystyle \left(x + \tfrac{5}{2}\right)^2 + (y-2)^2 \;=\;1\)


This is a circle: .\(\displaystyle \text{Center: }\left(-\tfrac{5}{2},\:2\right),\;\text{ radius } 1\) <<<<<<<< typo


There are no \(\displaystyle x\)-intercepts.

There are no \(\displaystyle y\)-intercepts.

A circle does not have vertices.

\(\displaystyle \text{Center: }\:\left(-\tfrac{5}{2},\:2\right)\)

\(\displaystyle \text{Domain: }\;x \in \left[-\tfrac{7}{2},\:-\tfrac{3}{2}\right] \)

\(\displaystyle \text{Range: }\;y \in[1,\:3] \)

... I don't want to pick at you but there is a tiny typo.