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.

x28y+y2+11=0x^2-8 y+y^2+11=0

Solution

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

x28y+y2+11=0\begin{aligned} x^2-8y+y^2+11=0 \end{aligned}

Use APE

x28y+y2+11+(11)=0+(11)\begin{aligned} x^2-8y+y^2+11+(-11)=0+(-11) \end{aligned}

Arrange

x2+[y28y]=11\begin{aligned} x^2+\left[y^2-8y\right]=-11 \end{aligned}

Complete the square

x2+[y28y+(b2)2]=11+(b2)2\begin{aligned} x^2+\left[y^2-8y+\left(\frac{b}{2}\right)^2\right]&=-11+\left(\frac{b}{2}\right)^2 \end{aligned}

Substitute

x2+[y28y+(82)2]=11+(82)2\begin{aligned} x^2+\left[y^2-8y+\left(\frac{-8}{2}\right)^2\right]&=-11+\left(\frac{-8}{2}\right)^2\end{aligned}

Simplify

x2+[y28y+(4)2]=11+(4)2x2+(y28y+16)=5\begin{aligned} x^2+\left[y^2-8y+(-4)^2\right]&=-11+(-4)^2\\ x^2+(y^2-8y+16)&=5 \end{aligned}

Factor

x2+(y4)2=5\begin{aligned} x^2+(y-4)^2=5 \end{aligned}

Hence the standard form is x2+(y4)2=5x^2+(y-4)^2=5 and indicates that the graph is Circle.

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