convert the equation to the standard form for an ellipse by completing the square on x and y

16x^2 + 25y^2 -32x -150y -159 = 0

Find the standard form of the equation of the hyperbola satisfying the given conditions:

center (4,7) Focus: (-3,7) vertex (3,7)




thanks guys could you explain these please?

x^2 - 6x + 8

is supposed to help me with

16x^2 + 25y^2 -32x -150y -159 = 0

if I was good at math I wouldn't be posting..

perhaps this example will help a bit more?

4x^2 + 9y^2 – 48x + 72y + 144 = 0

http://www.purplemath.com/modules/sqrellps.htm

thank you

anyone mind working the first one out? I'm getting wrong answer and following directions step by step..

Factorize the leading coefficients of X and Y :

16.X² - 32.X + 25.Y² - 150.Y - 159 = 0

16.( X² - 2.X) + 25.( Y² - 6.Y) - 159 = 0

Complete the square on X and Y (this can be done separately) :

16.( (X - 1)² - 1) + 25.( (Y - 3)² - 9) - 159 = 0

Expand a little :

16.( X - 1 )² - 16 + 25.( Y - 3 )² - 225 - 159 = 0

16.( X - 1 )² + 25.( Y - 3 )² = 16 + 225 + 159 = 400 = 20²

Divide both sides by 20² :

16.( X - 1 )² / 20² + 25.( Y - 3 )² / 20² = 20² / 20² = 1

Notice that 16 = 4² and that 25 = 5² :

4².( X - 1 )² / 20² + 5².( Y - 3 )² / 20² = 1

[ 4.( X - 1 ) / 20 ]² + [ 5.( Y - 3 ) / 20 ]² = 1

Cancel out some common factors :

[ ( X - 1 ) / 5 ]² + [ ( Y - 3 ) / 4 ]² = 1