Question

For the equation 16x^2 + 25y^2 -96x - 200y = -144, how do I write this in standard form? This is an ellipse.

Answer 1

@jim_thompson5910 @Directrix @Nnesha @perl

Answer 2

i got stuck and got 16(x-3)^2 + 25(y-2)^2 = 100. What do I do next?

Answer 3

@Hero @e.mccormick @freckles

Answer 4

first isn't 200/25=8 ?
\[16(x^2-6x)+25(y^2-?y)=-144 \\ 16(x^2-6x+9)+25(y^2-?y+?)=-144+16(9)+25(?) \\ \]

Answer 5

oh yeah, lolz my bad thnx @freckles

Answer 6

It will end up looking somethng like:

\(\dfrac{(x+h)^2}{a^2}+\dfrac{(y+k)^2}{b^2}=1\)

Answer 7

\[16(x^2-6x)+25(y^2-8y)=-144 \\ 16(x^2-6x+9)+25(y^2-8y+?)=-144+16(9)+25(?)\]
so what would that one question mark need to be to complete the square for the y part

Answer 8

16

Answer 9

so you add 25(16) on one side so we need to do that to the other
actually already put that one 25( ) over there
just left that other spot blank for the 16

Answer 10

\[16(x^2-6x+9)+25(y^2-8y+16)=-144+16(9)+25(16)\]

Answer 11

ok i got ((x-3)^2/25) + ((y-2)^2/16)

Answer 12

=1

Answer 13

but that 2 should be 4 ?

Answer 14

oh yeah forgot to fix that

Answer 15

thnx @freckles