L laladida88 New member Joined Jun 10, 2013 Messages 3 Jun 10, 2013 #1 3. Put the equation in standard from for an ellipse. 25x^2 +150x +9y^2 -8y =-9 I keep getting 1/4 (x +3) ^2 + 1/25 (y -2) ^2 =1 ? proof?
3. Put the equation in standard from for an ellipse. 25x^2 +150x +9y^2 -8y =-9 I keep getting 1/4 (x +3) ^2 + 1/25 (y -2) ^2 =1 ? proof?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jun 10, 2013 #2 Hello, laladida88! I assume there is a typo in the equation. 3\(\displaystyle \text{3. Write in standard form: }\:25x^2 + 150x + 9y^2 - \color{green}{18}y \:=\:-9\) Click to expand... We have: .. \(\displaystyle 25(x^2 + 6x \qquad) + 9(y^2 - 2y \qquad) \;=\;\text{-}9\) Complete the square: . . \(\displaystyle \color{red}{25}(x^2 + 6x \color{red}{+ 9}) + \color{blue}{9}(y^2 - 2y \color{blue}{+ 1}) \;=\;\text{-}9 \color{red}{+ 225} \color{blue}{+ 9}\) Factor: . . \(\displaystyle 25(x+3)^2 + 9(y-1)^2 \;=\;225\) Divide by 225: . . \(\displaystyle \dfrac{(x+3)^2}{9} + \dfrac{(y-1)^2}{25} \;=\;1\)
Hello, laladida88! I assume there is a typo in the equation. 3\(\displaystyle \text{3. Write in standard form: }\:25x^2 + 150x + 9y^2 - \color{green}{18}y \:=\:-9\) Click to expand... We have: .. \(\displaystyle 25(x^2 + 6x \qquad) + 9(y^2 - 2y \qquad) \;=\;\text{-}9\) Complete the square: . . \(\displaystyle \color{red}{25}(x^2 + 6x \color{red}{+ 9}) + \color{blue}{9}(y^2 - 2y \color{blue}{+ 1}) \;=\;\text{-}9 \color{red}{+ 225} \color{blue}{+ 9}\) Factor: . . \(\displaystyle 25(x+3)^2 + 9(y-1)^2 \;=\;225\) Divide by 225: . . \(\displaystyle \dfrac{(x+3)^2}{9} + \dfrac{(y-1)^2}{25} \;=\;1\)
