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Section 9.3 Conic Sections: Ellipses

Subsection 1. Complete the square

To graph a conic section, we complete the square in each variable to put the equation in standard form.

Subsubsection Example

Example 9.31.

Solve \(~2x^2+3 = 8x~\) by completing the square.

Solution

We begin by isolating the constant term on the right side of the equation.

\begin{align*} 2x^2 - 8x \amp =-3 \amp \amp \blert{\text{Factor out the coefficient of} ~x^2.}\\ 2(x+2-4x~~~~~~ \amp = -3) \amp \amp \blert{\text{Complete the square.}}\\ 2(x^2-4x \blert{+ 4}) \amp = -3 \blert{+ 4} \amp \amp \blert{\text{Simplify each side.}}\\ 2(x-2)^2 \amp = 5 \amp \amp \blert{\text{Isolate the perfect square.}}\\ (x-2)^2 \amp = \dfrac{5}{2} \amp \amp \blert{\text{Extract roots.}}\\ x-2 \amp = \pm \sqrt{\dfrac{5}{2}} \amp \amp \blert{\text{Sovle for}~x.} \end{align*}

Subsubsection Exercises

Solve \(~x^2-7x=4~\) by completing the square.

Answer
\(\dfrac{7 \pm \sqrt{65}}{2}\)

Solve \(~3x^2+6x-2=0~\) by completing the square.

Answer
\(-1 \pm \sqrt{\dfrac{5}{3}}\)

Solve \(~2x^2-8=3x~\) by completing the square.

Answer
\(\dfrac{3 \pm \sqrt{73}}{4}\)

Subsection 2. Find points on a graph

Points on the graph of a conic section satisfy a quadratic equation in two variables.

Subsubsection Example

Example 9.35.

Find all points on the graph of \(x^2+y^2 = 12\) with \(x\)-coordinate 2.

Solution

We substitute \(x=2\) into the equation to obtain

\begin{equation*} 2^2+y^2=12 \end{equation*}

and simplify to \(y^2=8\text{.}\) Solving for \(y\text{,}\) we find \(y=\pm \sqrt{8} = \pm 2\sqrt{2}\text{.}\) Thus, the points on the graph of \(x^2+y^2 = 12\) with \(x\)-coordinate 2 are \((2, 2\sqrt{2})\) and \((2, -2\sqrt{2})\text{.}\)

Subsubsection Exercises

Find all points on the graph of \(y=2x^2-4x+3\) with \(y\)-coordinate 3.

Answer
\((0,3),~(2,3)\)

Find all points on the graph of \((x-4)^2+(y+1)^2=25\) with \(y\)-coordinate 3.

Answer
\((1,3)~(7,3)\)

Find all points on the graph of \(\dfrac{(x-2)^2}{16}+\dfrac{(y+1)^2}{9} = 1\) with \(x\)-coordinate \(-1\text{.}\)

Answer
\(\left(-1,-1+\dfrac{3\sqrt{7}}{4}\right),~\left(-1,-1-\dfrac{3\sqrt{7}}{4}\right)\)

Subsection 3. Divide by a fraction

We may encounter coefficients that are fractions when putting anequation in standard form.

Subsubsection Example

Example 9.39.

Solve for \(y\text{:}\) \(~\dfrac{6}{5}y = \dfrac{2}{3}x^2-8x+12\)

Solution

We divide both sides of the equation by \(~\dfrac{6}{5}\text{,}\) or equivalently, we multiply both sides by \(~\dfrac{5}{6}\text{.}\)

\begin{align*} \blert{\dfrac{5}{6}} \left(\dfrac{6}{5}y\right) \amp = \blert{\dfrac{5}{6}} \left(\dfrac{2}{3}x^2-8x+12\right) \amp \amp \blert{\text{Apply the distributive law.}}\\ y \amp = \blert{\dfrac{5}{6}} \left(\dfrac{2}{3}x^2\right)-\blert{\dfrac{5}{6}}(8x) + \blert{\dfrac{5}{6}}(12) \amp \amp \blert{\text{Simplify each term.}}\\ y \amp = \dfrac{5}{9}x^2-\dfrac{20}{3}x+10 \end{align*}

Subsubsection Exercises

Solve for \(y\text{:}\) \(~\dfrac{2}{3}y - (x+5)^2 = 1\)

Answer
\(y = \dfrac{3}{2}(x+5)^2 + \dfrac{3}{2}\)

Solve for \(y\text{:}\) \(~2(x-6)^2 = \dfrac{4}{5}y\)

Answer
\(y = \dfrac{5}{2}(x-6)^2\)

Solve for \(z\text{:}\) \(~2 (x+1)^2 + 6(y-w)^2 = \dfrac{24}{9}z\)

Answer
\(z = \dfrac{3}{4}(x+1)^2 + \dfrac{9}{4}(y-3)^2\)