Question

etermine if $9x^2 + 16y^2 + 36x + 128y + 148 = 0$ is an ellipse.
\bigcirc not an ellipse
\bigcirc ellipse, we can write the equation of the ellipse in standard form $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.
where:
$h = \square$
$k = \square$
$a = \square$
$b = \square$
question help: \video video 1 \video video 2 \message instructor
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Explanation:

Step1: Group and factor \(x\) and \(y\) terms

Group \(x\) - terms and \(y\) - terms:
\(9x^{2}+36x + 16y^{2}+128y=-148\)
Factor out the coefficients of \(x^{2}\) and \(y^{2}\):
\(9(x^{2}+4x)+16(y^{2}+8y)=-148\)

Step2: Complete the square for \(x\) and \(y\)

For the \(x\) - part: \((x + 2)^{2}=x^{2}+4x + 4\), so \(x^{2}+4x=(x + 2)^{2}-4\)
For the \(y\) - part: \((y + 4)^{2}=y^{2}+8y + 16\), so \(y^{2}+8y=(y + 4)^{2}-16\)

Substitute back into the equation:
\(9[(x + 2)^{2}-4]+16[(y + 4)^{2}-16]=-148\)
\(9(x + 2)^{2}-36+16(y + 4)^{2}-256=-148\)
\(9(x + 2)^{2}+16(y + 4)^{2}=-148 + 36+256\)
\(9(x + 2)^{2}+16(y + 4)^{2}=144\)

Step3: Divide by 144 to get the standard form

Divide each term by 144:
\(\frac{(x + 2)^{2}}{16}+\frac{(y + 4)^{2}}{9}=1\)

Answer:

Ellipse, we can write the equation of the ellipse in standard form \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\). Where:
\(h=-2\)
\(k=-4\)
\(a = 4\)
\(b = 3\)