## Related questions with answers

Write the equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation.

$x^2-8 y+y^2+11=0$

Solution

VerifiedRewrite the given into standard form.

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

Use APE

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

Arrange

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

Complete the square

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

Substitute

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

Simplify

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

Factor

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

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

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