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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.

3x2+4y2+8y=83 x^2+4 y^2+8 y=8

Solution

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Answered 2 years ago
Answered 2 years ago
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Rewrite the given into standard form.

3x2+4y2+8y=8\begin{aligned} 3x^2+4y^2+8y=8 \end{aligned}

Factor out 44

3x2+4[y2+2y]=8\begin{aligned} 3x^2+4\left[y^2+2y\right]=8 \end{aligned}

Complete the square

3x2+4[y2+2y+(b2)2]=8+4(b2)2\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

3x2+4[y2+2y+(22)2]=8+4(22)2\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

3x2+4[y2+2y+(1)2]=8+4(1)23x2+4(y2+2y+1)=12\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

3x2+4(y+1)2=12\begin{aligned} 3x^2+4(y+1)^2=12 \end{aligned}

Equate to 11

3x212+4(y+1)212=1212x24+(y+1)23=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 x24+(y+1)23=1\dfrac{x^2}{4}+\dfrac{(y+1)^2}{3}=1 and indicates that the graph is Ellipse.

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