Question

Write in standard form of parabola:
1. 9x^2+36x+25y^2-50y=164
2. x^2+y^2-10x+4y+27=0
3. 45y^2-320x^2+6=2886

Answer 1

Do you have any ideas on this?

Answer 2

@IMStuck This would basically be simplifying it right?

Answer 3

no
since \( x^2+y^2-10x+4y+27=0\) is not a parabola!

Answer 4

actually, none are parabolas!

Answer 5

Well I just have to write all of them in standard form starting from finding the vertex

Answer 6

Wow @satellite73 , you are clearly very knowledgeable :)

Answer 7

you cannot find the vertex
they are not parabolas

Answer 8

Ok well how do i write these in standard form?

Answer 9

\[9x^2+36x+25y^2-50y=164\]is an ellipse

Answer 10

I was going to say that these are not parabolas at all. The first one could be a hyperbola. It most likely is!

Answer 11

you can write it in standard form of an ellipse if you like
you will end up with\[\frac{(xh)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]

Answer 12

No forget that...ellipse!

Answer 13

Ok I am still very confused... can you demonstrate one of them?

Answer 14

\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]rather
it is an annoying piece of algebra called "completing the square"

Answer 15

i personally like completing the square! No?

Answer 16

i can walk you through the first one if you like \[9x^2+36x+25y^2-50y=164\]start by factoring out the common factor of the \(x\) and \(y\) terms and start with
\[9(x^2+4x)+25(y^2-2y)=164\]

Answer 17

then take half of the coefficient of the linear term, write as a complete square via
\[25(x+2)^2+25(y-1)^2=164+9\times 2^2+25\times 1^2\]

Answer 18

You are really doing a good job for madisonhope96. Yay for you. These are not the easiest things to get through!

Answer 19

ok i follow you

Answer 20

damn typo!!\[9(x+2)^2+25(y-1)^2=164+9\times 2^2+25\times 1^2\]

Answer 21

Ok, I still got what you meant from the first demonstration! :)

Answer 22

arithmetic on the right gives
\[9(x+2)^2+25(y-1)^2=225\]

Answer 23

divide by \(225\) and get \[\frac{(x+1)^2}{25}+\frac{(y-1)^2}{9}=1\]

Answer 24

which looks like \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]exactly

Answer 25

I'm getting that the second one is a circle.

Answer 26

Thank you so much! Now how would you find the minor vertices of that problem if you don't mind me asking?

Answer 27

yet another typo!\[\frac{(x+2)^2}{25}+\frac{(y-1)^2}{9}=1\]

Answer 28

center is \((-2,1)\)

Answer 29

So that is the only minor vertices? Im so confused sorry

Answer 30

minor axis vertices*

Answer 31

i have never heard of "minor axis vertices"

Answer 32

Ok thank you!

Answer 33

if you want the length of the major axis, it is \(10\) and the length of the minor axis is \(6\)

Answer 34

\[\frac{(x+2)^2}{25}+\frac{(y-1)^2}{9}=1\]\[\frac{(x+2)^2}{5^2}+\frac{(y-1)^2}{3^2}=1\]\[a=5,b=3\]

Answer 35

The minor axis vertices lie on the axis that the ellipse is NOT centered about. In other words, if the x axis is the major vertice axis and the foci lie on this axis, then the other axis is the minor. In an ellipse, a is ALWAYS greater than b, and a = 5 and b = 3. The 5 is under the x, so the x is the major vertice and y is the minor.

Answer 36

actually although it will not tell you how to get it, any information about this ellipse can be found here
http://www.wolframalpha.com/input/?i=ellipse+%28x%2B2%29^2%2F25%2B%28y-1%29^2%2F9%3D1&dataset=&equal=Submit

Answer 37

The coordinates for the major vertices are (3, 1) and (-7, 1) and the coordinates for the minor vertices are (-2, 4) and (-2, -2)

Answer 38

Thank you so much!!!

Answer 39

Do you understand how to find the major and minor vertices? It's quite simple, just takes some explaining.

Answer 40

Yes, that clarified a lot! Now the last thing I'm confused on is how to find the value of p for -2x^2+16x+24y-224=0???

Answer 41

Ok so if you have a p to find, this tells me right away that this is a parabola.

Answer 42

Ok, so I'm just confused as to where exactly I would start?

Answer 43

You must complete the square in this one too. The standard form for a parabola is x^2=4py, where p is the distance that the focal point is from the vertex.

Answer 44

i will do step by step as did mr. satellite73. Ok?

Answer 45

Ok thank you!

Answer 46

\[-2x ^{2}+16x=-24y+224\]Or\[2x ^{2}-16x=24y-224\]Complete the square on the x terms (this is an x^2 parabola, meaning it opens either upwards or downwards. I tell my students it is either a cup or a mountain).

Answer 47

\[2(x ^{2}-8x)=24y-224\]\[2(x ^{2}-8x+16)=24y-224+2(16)\]\[2(x-4)^{2}=24y-192\]\[(x-4)^{2}=12y-96\]\[(x-4)^{2}=12(y-8)\]

Answer 48

Ok I understand that better now

Answer 49

Now here is the point at which you are able to solve for p. The method used for this is 4p=coefficient of y. Which is 4p = 12, so p = 3. That means that the focus is 3 units up from the vertex of (4, 8) and it is a "cup" parabola, because the focus is above the vertex. That's the whole equation; you kind of needed to do the whole thing in order to find p.

Answer 50

These are not easy; it was THE hardest chapter for my Algebra 2 kids this past year. Keep at it, memorize the rules for each type of quadratic equation and it will make it easier for you. Promise.

Answer 51

Oh! I originally got that, but thought I needed to do more at first! So p=3 is all i need? Thank you sooooo much!

Answer 52

You've most very welcome!!!