Question

question 14 · 1 point
find the standard form of the ellipse given below in general form.
36x² + 144x + 25y² + 250y - 131 = 0

Explanation:

Step1: Complete the square for x - terms

Group the x - terms: $36x^{2}+144x=36(x^{2}+4x)$.
Completing the square inside the parentheses: $x^{2}+4x=(x + 2)^{2}-4$. So, $36(x^{2}+4x)=36((x + 2)^{2}-4)=36(x + 2)^{2}-144$.

Step2: Complete the square for y - terms

Group the y - terms: $25y^{2}+250y=25(y^{2}+10y)$.
Completing the square inside the parentheses: $y^{2}+10y=(y + 5)^{2}-25$. So, $25(y^{2}+10y)=25((y + 5)^{2}-25)=25(y + 5)^{2}-625$.

Step3: Rewrite the original equation

The original equation $36x^{2}+144x + 25y^{2}+250y-131 = 0$ becomes:
$36(x + 2)^{2}-144+25(y + 5)^{2}-625-131 = 0$.

Step4: Simplify the equation

$36(x + 2)^{2}+25(y + 5)^{2}=144 + 625+131$.
$36(x + 2)^{2}+25(y + 5)^{2}=900$.

Step5: Convert to standard form

Divide both sides by 900: $\frac{36(x + 2)^{2}}{900}+\frac{25(y + 5)^{2}}{900}=1$.
Which simplifies to $\frac{(x + 2)^2}{25}+\frac{(y + 5)^2}{36}=1$.

Answer:

$\frac{(x + 2)^2}{25}+\frac{(y + 5)^2}{36}=1$