SOLUTION: Write the equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola .
x2 + 4y2 + 2x - 24y + 33 = 0
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-> SOLUTION: Write the equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola .
x2 + 4y2 + 2x - 24y + 33 = 0
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Question 635404: Write the equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola .
x2 + 4y2 + 2x - 24y + 33 = 0 Found 2 solutions by ewatrrr, AnlytcPhil:Answer by ewatrrr(24785) (Show Source):
Hi,
x2 + 4y2 + 2x - 24y + 33 = 0
(x+1)^2 + 4(y-3)^2 - 1 - 36 + 33 = 0
(x+1)^2 + 4(y-3)^2 = 4
Ellipse: C(-1,3)
Standard Form of an Equation of an Ellipse is
where Pt(h,k) is the center. (a variable positioned to correspond with major axis)
a and b are the respective vertices distances from center
and ±are the foci distances from center: a > b
x² + 4y² + 2x - 24y + 33 = 0
Rearrange equation like this
(x² + 2x) + (4y² - 24y) = -33
(x² + 2x) + 4(y² - 6y) = -33
Complete the square:
Multiply the coefficient of each 1st degree term
by one-half theb square the result. Then add this
inside each set of parentheses, and add the
corresponding amount to the right side:
(x² + 2x + 1) + 4(y² - 6y + 9) = -33 + 1 + 36
Note that since 9 was added in the second set of parenthese,
that ammounted to adding 36 to both side because of 4 coefficient
of the second set of parentheses.
Factor the quadratics inside the parentheses as perfect squarse,
and combine terms on the right side:
(x+1)² + 4(y-3)² = 4
Get 1 on the right side by dividing every term though by 4
= = =
Edwin
