Question

16x2 + 25y2 - 96x - 200y + 144 = 0....can someone please help me put this in standard form?

Answer 1

ellipse right?

Answer 2

yes!

Answer 3

i dont think so....

Answer 4

answer or method? method is a pain, answer is easy
you are going to have to complete the square to make it look like
\[16(x-h)^2+25(y-k)^2=C\] and then divide to get
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]

Answer 5

\[16x^2 + 25y^2 - 96x - 200y + 144 = 0\]
\[16(x^2-6x)+25(y^2-4)=-144\] is your start

Answer 6

then to complete the square take half of 6, square it to get 9 and multiply by 16 to get 144
also take half of 4, which is 2, square to get 4, multiply by 25 to get 100
and write
\[16(x-3)^2+25(y-2)^2=-144+144+100\]

Answer 7

oh damn i make a mistake!!

Answer 8

\[16(x^2-6x)+25(y^2-4)=-144\] should be
\[16(x^2-6x)+25(y^2-8)=-144\]

Answer 9

ok im following so far..this is great

Answer 10

you could just do answer if you want

Answer 11

ok same idea, this time get
\[16(x-3)^2+25(y-4)^2=-144+144+400\]

Answer 12

i made a mistake it should have been 8 instead of 4
therefore half of 8 is 4, 4 squares is 16 and 16 times 25 is 400

Answer 13

so you get
\[16(x-3)^2+25(y-4)^2=400\]

Answer 14

your last step is to divide both sides by 400 to put a 1 on the right, and you are done

Answer 15

sorry but what would that look like?

Answer 16

what is \(\frac{16}{400}\)?

Answer 17

1/25

Answer 18

so first term is \(\frac{(x-3)^2}{25}\)

Answer 19

ok got it all

Answer 20

THANKS!

Answer 21

do you know how to find the foci of this?

Answer 22

the center is (3,4)

Answer 23

@satellite73