Expert: Socrates - 5/26/2004

Question
classify the graph as a parabola, circle, ellipse, or hyperbola. then write the equation in standard form.

1) 4x^2+ y=3
2) 3x^2+6x-6y^2+24y-33=0


Answer
1)

4x^2 + y = 3

y = -4x^2 + 3

You should recognize that the graph of this equation is a parabola.

The standard form is (x-h)^2 = 4p(y-k). So we need to get the given equation into this form.

4x^2 + y = 3

4x^2 = -y + 3

x^2 = (1/4) (-y + 3)

x^2 = (-1/4) (y - 3)

x^2 = (4)(-1/16) (y - 3)

(x - 0 )^2 = (4)(-1/16) (y - 3)

the last equation is the standard form with
h=0 , p=-1/16 , k=3


2)

3x^2 + 6x - 6y^2 + 24y - 33 = 0

divide through by 3

x^2 + 2x - 2y^2 + 8y - 11 = 0

then

x^2 + 2x - 2y^2 + 8y = 11

complete the square in x by adding 1 to both sides

x^2 + 2x + 1 - 2y^2 + 8y = 12

(x+1)^2 - 2y^2 + 8y = 12

(x+1)^2 - (2) (y^2 - 4y) = 12

complete the square in y by subtracting 8 from both sides

(x+1)^2 - (2) (y^2 - 4y + 4) = 4

(x+1)^2 - (2) (y-2)^2 = 4

divide through by 4

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

You should now recognize this as the standard form of the equation of the hyperbola,

(x-h)^2 / a^2  -  (y-k)^2 / b^2 = 1

with h=-1 , k=2 , a=2 , b=2^1/2