# 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

Algebra ->  Quadratic-relations-and-conic-sections -> 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      Log On

 Click here to see ALL problems on Quadratic-relations-and-conic-sections 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(22107)   (Show Source): You can put this solution on YOUR website! ``` 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 ```Answer by AnlytcPhil(1539)   (Show Source): You can put this solution on YOUR website! ```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```