Question

write the equation of the ellipse ( 9 x^{2}+36 y^{2}-54 x + 288 y + 333 = 0 ) in standard form

\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}} = 1,
where:
( h = )
( k = )
( a = )
( b = )
question help: video 1 video 2 written example 1

Explanation:

Step1: Group x and y terms

Group the \(x\)-terms and \(y\)-terms together:
\(9x^2 - 54x + 36y^2 + 288y = -333\)

Step2: Factor out coefficients of squared terms

Factor out 9 from the \(x\)-terms and 36 from the \(y\)-terms:
\(9(x^2 - 6x) + 36(y^2 + 8y) = -333\)

Step3: Complete the square for x

For \(x^2 - 6x\), take half of -6 (\(-3\)), square it (\(9\)):
\(9(x^2 - 6x + 9) + 36(y^2 + 8y) = -333 + 9 \times 9\)

Step4: Complete the square for y

For \(y^2 + 8y\), take half of 8 (\(4\)), square it (\(16\)):
\(9(x^2 - 6x + 9) + 36(y^2 + 8y + 16) = -333 + 81 + 36 \times 16\)

Step5: Simplify both sides

Simplify the right-hand side:
\(9(x - 3)^2 + 36(y + 4)^2 = -333 + 81 + 576\)
\(9(x - 3)^2 + 36(y + 4)^2 = 324\)

Step6: Divide by 324 to get standard form

Divide each term by 324:
\(\frac{9(x - 3)^2}{324} + \frac{36(y + 4)^2}{324} = 1\)
\(\frac{(x - 3)^2}{36} + \frac{(y + 4)^2}{9} = 1\)

Answer:

\(h = 3\), \(k = -4\), \(a = 6\), \(b = 3\)