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Question

# Write the equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation.$3 x^2+4 y^2+8 y=8$

Solution

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Step 1
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Rewrite the given into standard form.

\begin{aligned} 3x^2+4y^2+8y=8 \end{aligned}

Factor out $4$

\begin{aligned} 3x^2+4\left[y^2+2y\right]=8 \end{aligned}

Complete the square

\begin{aligned} 3x^2+4\left[y^2+2y+\left(\frac{b}{2}\right)^2\right]=8+4\left(\frac{b}{2}\right)^2 \end{aligned}

Substitute

\begin{aligned} 3x^2+4\left[y^2+2y+\left(\frac{2}{2}\right)^2\right]=8+4\left(\frac{2}{2}\right)^2 \end{aligned}

Simplify

\begin{aligned} 3x^2+4\left[y^2+2y+(1)^2\right]&=8+4(1)^2\\ 3x^2+4(y^2+2y+1)&=12 \end{aligned}

Factor

\begin{aligned} 3x^2+4(y+1)^2=12 \end{aligned}

Equate to $1$

\begin{aligned} \frac{3x^2}{12}+\frac{4(y+1)^2}{12}=\frac{12}{12}\\ \frac{x^2}{4}+\frac{(y+1)^2}{3}&=1 \end{aligned}

Hence, the standard form is $\dfrac{x^2}{4}+\dfrac{(y+1)^2}{3}=1$ and indicates that the graph is Ellipse.

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