Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices and lengths of the major and minor axes. If it is a parabola, find the vertex, focus and directrix. If it is a hyperbola, find the center, foci, vertices and asymptotes. Sketch the graph of the equation. If the equation has no graph, explain why.
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\begin{equation}
\begin{aligned}
9x^2- 36x + 4y^2 =& 0
&& \text{Factor and group terms}
\\
\\
9(x^2 - 4x + ) + 4y^2 =& 0
&& \text{Complete the square: add } \left( \frac{4}{2} \right)^2 = 4 \text{ on the right side and 36 on the left side}
\\
\\
9(x^2 - 4x + 4) + 4y^2 =& 36
&& \text{Complete the square}
\\
\\
9(x - 2)^2 + 4y^2 =& 36
&& \text{Divide by } 36
\\
\\
\frac{(x - 2)^2}{4} + \frac{y^2}{9} =& 1
&&
\end{aligned}
\end{equation}
$
We can say that the equation is an ellipse since it is the sum of the squares. The equation $\displaystyle \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$ has a vertical major axis and vertices on $(0, \pm a)$, center on $(h, k)$ and foci $(0, \pm c)$ where $c = \sqrt{a^2 - b^2}$.
Also, the length of its major and minor axis is $2a$ and $2b$ respectively, So if $a^2 = 9$ and $b^2 = 4$, then $a = 3, b = 2$ and $c = \sqrt{9 - 4} = \sqrt{5}$.
Thus, by applying transformations,
center $(2,0)$
vertices $(2, 0 \pm 3) \to (2, \pm 3)$
foci $(2, 0 \pm \sqrt{5}) \to (2, \pm \sqrt{5})$
length of major axis $6$
length of minor axis $4$
Therefore, the graphs is
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