Question

given 5x^2+9y^2-150x-18y+9=0 , how to make this equation equal to 1? this is in elipse

Answer 1

Well I think you mean you want to put it in this form:
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]

Answer 2

We need to complete the square..
Once for the x part and another time for the y part
x^2+bx+(b/2)^2=(x+b/2)^2

Answer 3

\[5x^2-150x+9y^2-18y=-9 \\ 5(x^2-30x)+9(y^2-2y)=-9 \\ 5(x^2-30x+?_1)+9(y^2-2y+?_2)=-9+5?_1+9?_2\]

Answer 4

See if you can figure out what those ? marks need to be

Answer 5

okay i'm trying, wait a sec

Answer 6

sorry i'm not clear. whats for '?'

Answer 7

is it 1?

Answer 8

We need to complete the square for both x part and y part

Answer 9

I will do x part but you can do y part

Answer 10

\[5x^2-150x+9y^2-18y=-9 \\ 5(x^2-30x)+9(y^2-2y)=-9 \\ 5(x^2-30x+?_1)+9(y^2-2y+?_2)=-9+5?_1+9?_2 \\ 5(x^2-30x+15^2)+9(y^2-2y+?_2)=-9+5(15^2)+9?_2 \\ 5(x-15)^2+9(y^2-2y+?_2)=-9+5(15^2)+9?_2\]

Answer 11

So you need to do the y part now
remember you are trying to complete the square

Answer 12

okay

Answer 13

why use 15^2?

Answer 14

Because I'm using it to complete the square..
Did you read my second post

Answer 15

x^2+bx+(b/2)^2=(x+b/2)^2

Answer 16

adding in (b/2)^2 allows us to write it as something squared

Answer 17

what is the value of b?

Answer 18

ohh i got it

Answer 19

for y ; 9(y^2 - 2y+1)
5(x-15)^2 + 9(y^2 -2y+1) = -9+5(15)^2 + 9(1^2) ?

Answer 20

\[5(x-15)^2+9(y-1)^2=-9+5(15^2)+9(1^2) \\ 5(x-15)^2+9(y-1)^2=5(15^2)\]

Answer 21

Now what is 15^2*5
225*5
---
1125
I think that is 1125...You might want to check that.

Answer 22

\[5(x-15)^2+9(y-1)^2=1125\]
you wanted 1 on the right hand side
so divide both sides by 1125

Answer 23

ohh i see.. how (x^2 - 30x +15^2) become (x-15)^2 ? need to simplify?

Answer 24

Just completing the square
(and I just want to rewrite my formula with a | | on that b/2)...
\[x^2+bx+(\frac{b}{2})^2=(x+|\frac{b}{2}|)^2 \\ x^2-bx+(\frac{b}{2})^2=(x-|\frac{b}{2}|)^2\]

Answer 25

\[x^2+100x+(\frac{100}{2})^2=(x+\frac{100}{2})^2 \\\ x^2-100x+(\frac{100}{2})^2=(x-\frac{100}{2})^2\]

Answer 26

\[ax^2+by^2+cx+dy+e=0 \\ ax^2+cx+by^2+dy=-e \\ ax^2+\frac{a}{a}cx+by^2+\frac{b}{b}dy=-e \\ (ax^2+\frac{a}{a}cx)+(by^2+\frac{b}{b}dy)=-e \\ a(x^2+\frac{1}{a}cx)+b(y^2+\frac{1}{b}dy)=-e \\ a(x^2+\frac{c}{a}x)+b(y^2+\frac{d}{b}y)=-e \\ a(x^2+\frac{c}{a}x+?_1)+b(y^2+\frac{d}{b}y+?_2)=-e+a?_1+b?_2 \\ a(x^2+\frac{c}{a}x+(\frac{c}{2a})^2)+b(y^2+\frac{d}{b}y +(\frac{d}{2b})^2=-e+a(\frac{c}{2a})^2+b(\frac{d}{2b})^2 \\ a(x+\frac{c}{2a})^2+b(y+\frac{d}{2b})^2=-e+\frac{ac^2}{4a^2}+\frac{bd^2}{4b^2} \\ a(x+\frac{c}{2a})^2+b(y+\frac{d}{2b})^2=-e+\frac{c^2}{4a}+\frac{d^2}{4b} \\a(x+\frac{c}{2a})^2+b(y+\frac{d}{2b})^2=\frac{-4eab+c^2b+d^2a}{4ab}\]

Answer 27

i see..thanks for remind again. i learnt completing the square 4 years ago and i forgot already :D

Answer 28

\[\frac{a(x+\frac{c}{2a})^2}{\frac{-4eab+c^2b+d^2a}{4ab}}+\frac{b(y+\frac{d}{2b})^2}{\frac{-4eab+c^2b+d^2a}{4ab}}=1\]

Answer 29

lol If I didn't make a mistake...

Answer 30

If I did this correctly, you should able to plug in your numbers from 5x^2+9y^2-150x-18y+9=0 and get the same thing we already got

Answer 31

a=5,b=9,c=-150, d=-18,e=9

Answer 32

\[\frac{5(x+\frac{-150}{2(5)})^2}{\frac{-4(9)(5)(9)+(-150)^2(9)+(-18)^2(5)}{4(5)(9)}}+\frac{9(y+\frac{-18}{2(9)})^2}{\frac{-4(9)(5)(9)+(-150)^2(9)+(-18)^2(5)}{4(5)(9)}}=1 \\ \frac{5(x+-15)^2}{1125}+\frac{9(y+-1)^2}{1125}=1 \]

Answer 33

http://www.wolframalpha.com/input/?i=%28-4%289%29%285%29%289%29%2B%28-150%29%5E2%289%29%2B%28-18%29%5E2%285%29%29%2F%284%285%29%289%29%29
I didn't fill like doing the calculations in the bottom by hand :p

Answer 34

As you a formula for this is really ugly

Answer 35

i see.. thanks a lot for help and guide. i really no idea to solve. in class, the question really simple and easy.when doing task, its so challenging. ;)