Question

16x^2+128x+81y^2-324y-716=0
Write the equation in the standard form of an ellipse with center (h, k)

Answer 1

lol oh i see
http://www.sosmath.com/algebra/quadraticeq/complsquare/complsquare.html

Answer 2

Complete the square.

Answer 3

16x^2+128x+81y^2-324y=716

Answer 4

ok so were trying to get it in the form (x-h)^2+(y-k)^2=r^2

Answer 5

you sure thats suppose to be 81y^2?

Answer 6

Well, it is important to get the exponent of x^2 and y^2 to 1. So first change to this form\[16(x ^{2}+8x)+81(y ^{2}-4)=716\]

Answer 7

yes...

Answer 8

\[81(y ^{2}-4x)\]

Answer 9

-4y

Answer 10

then we get?

16(x^2+8x+16)+81(y^2-4y+4)=736

16(x+4)^2+81(y-2)^2=736

Answer 11

so from the general form of the equation (x-h)^2+(y-k)^2=r^2

our center (h,k) is (-4,2)

Answer 12

and you can check it by the general form of a circle which is

Ax^2+Ay^2+Bx+Cy+D=0

and the formula for the center is (-b/2a,-c/2a)

Answer 13

if you have any question feel free to ask but check out the link i first entered thats what helped me

Answer 14

Wait, wait. Ellipse, we like the RHS to be 1, so divide everything by 736.

Answer 15

oh okay yes sorry

Answer 16

Good work, purplec. You're a stud.

Answer 17

thank you...and stud?..lol

Answer 18

You're a rockstar!