Question

Change to standard form 2x^2+4y^2-12x-64y-16=0 i don't know how to solve it.

Answer 1

you will have to "complete the square"

2x^2+4y^2-12x-64y-16=0
2(x^2 - 6x + ....) + 4(y^2 - 16y + ....) = 16
2(x^2 - 6x + 9) + 4(y^2 - 16y + 64) = 16 + 2(9) + 4(64)
2(x-3)^2 + 4(y-8)^2 = 290
divide each term by 290
2(x-3)^2 /290 + 4(y-8)^2 / 290 = 1
(x-3)^2 / 145 + (y-8)^2 / (145/2) = 1

standard form:
(x-h)^2 /a^2 + (y-k)^2 /b^2 = 1 <----- ellipse with major axis as 2a, and minor axis as 2b, centre (h,k)

so centre is (3,8)
a = √145 , b = √(145/2)

Answer 2

2 x ^ 2 + 4 y ^ 2 - 12 x - 64 y - 16 = 0 Add 16 to both sides

2 x ^ 2 + 4 y ^ 2 - 12 x - 64 y - 16 + 16 = 0 + 16

2 x ^ 2 + 4 y ^ 2 - 12 x - 64 y = 16

2 x ^ 2 - 12 x + 4 y ^ 2 - 64 y = 16

2 ( x ^ 2 - 6 x ) + 4 ( y ^ 2 - 16 y ) = 16 Divide both sides by 4

2 ( x ^ 2 - 6 x ) / 4 + 4 ( y ^ 2 - 16 y ) / 4 = 16 / 4

( 2 / 4 ) ( x ^ 2 - 6 x ) + ( 4 / 4 ) ( y ^ 2 - 16 y ) = 4

( 1 / 2 ) ( x ^ 2 - 6 x ) + ( y ^ 2 - 16 y ) = 4 Add 9 / 2 to both sides

( 1 / 2 ) ( x ^ 2 - 6 x ) + ( y ^ 2 - 16 y ) + 9 / 2 = 4 + 9 / 2

( 1 / 2 ) ( x ^ 2 - 6 x ) + 9 / 2 + ( y ^ 2 - 16 y ) = 8 / 2 + 9 / 2

( 1 / 2 ) ( x ^ 2 - 6 x + 9 ) + ( y ^ 2 - 16 y ) = 17 / 2 Add 64 to both sides

( 1 / 2 ) ( x ^ 2 - 6 x + 9 ) + ( y ^ 2 - 16 y ) + 64 = 17 / 2 + 64

( 1 / 2 ) ( x ^ 2 - 6 x + 9 ) + ( y ^ 2 - 16 y + 64 ) = 17 / 2 + 128 / 2

( 1 / 2 ) ( x ^ 2 - 6 x + 9 ) + ( y ^ 2 - 16 y + 64 ) = 145 / 2


Since the ( x ^ 2 - 6 x + 9 ) = ( x - 3 ) ^ 2 and y ^ 2 - 16 y + 64 = ( y - 8 ) ^ 2


( 1 / 2 ) ( x - 3 ) ^ 2 + ( y - 8 ) ^ 2 = 145 / 2 Multiply both sides by 2

( 1 / 2 ) * 2 * ( x - 3 ) ^ 2 + 2 * ( y - 8 ) ^ 2 = 145 * 2 / 2

( x - 3 ) ^ 2 + 2 ( y - 8 ) ^ 2 = 145 Divide both sides by 145

( x - 3 ) ^ 2 / 145 + 2 ( y - 8 ) ^ 2 / 145 = 145 / 145

( 1 / 145 ) ( x - 3 ) ^ 2 + ( 2 / 145 ) ( y - 8 ) ^ 2 = 1