Question 1

Please write
$4 {x}^(2) + 9 {y}^(2) - 24x + 18y + 9 = 0$ in standard form

Question 2

$\begin{gathered} 4x^2+9y^2-24x+18y+9=0 \\ 4x^2-24x+9y^2+18y+9=0\text{ (Organizing)} \\ 4x^2-24x+9y^2+18y=-9\text{ (Subtracting -9 from both sides of the equation)} \\ 4(x^2-6x)+9(y^2+2y)=-9\text{ (Factoring)} \\ 4(x^2-6x+((-6)/(2))^2)+9(y^2+2y+((2)/(2))^2)=-9+4\cdot((-6)/(2))^2+9\cdot((2)/(2))^2\text{ (Completing the square)} \end{gathered}$
$\begin{gathered} 4(x^2-6x+9^{})+9(y^2+2y+1^{})=-9+36+9\text{ (Dividing and raising the result to the power of 2)} \\ 4\mleft(x^2-6x+9^{}\mright)+9\mleft(y^2+2y+1^{}\mright)=36\text{ (Subtracting)} \\ 4(x-3^{})^2+9(y+1^{})^2=36\text{ (Factoring)} \\ \frac{4(x-3^{})^2}{36}+\frac{9(y+1^{})^2}{36}=(36)/(36)\text{ (Dividing on both sides of the equation by 36)} \\ \frac{(x-3^{})^2}{9}+\frac{(y+1^{})^2}{4}=1\text{ (Simplifying)} \\ \text{The answer is }\frac{(x-3^{})^2}{9}+\frac{(y+1^{})^2}{4}=1 \end{gathered}$

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