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Section 9.4 Conic Sections: Hyperbolas

Subsection 1. Write a quadratic equation in standard form

The parameters in the standard form determine the shape of the graph.

Subsubsection Example

Example 9.43.

Write the equation \(~4x^2-3y^2=1~\) in the form \(~\dfrac{x^2}{A}-\dfrac{y^2}{B} = 1\)

Solution

Recall that dividing by a fraction is equivalent to multiplying by its reciprocal. For instance, \(~\dfrac{x^2}{\dfrac{1}{2}} = 2x~\text{,}\) and \(~\dfrac{a}{\dfrac{3}{4}} = \dfrac{4}{3}a~\)

Thus, for our example,

\begin{equation*} ~4x^2 = \dfrac{x^2}{\dfrac{1}{4}},~~~ \text{and} ~~~3y^2 = \dfrac{y^2}{\dfrac{1}{3}}~ \end{equation*}

So we can write the equation as \(~\dfrac{x^2}{\dfrac{1}{4}} - \dfrac{y^2}{\dfrac{1}{3}} = 1\text{.}\)

Subsubsection Exercises

Write each equation in the form \(~\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1\)

\(\dfrac{4}{9}(x-2)^2 - \dfrac{1}{5}(y+2)^2 = 1\)

Answer
\(\dfrac{(x-2)^2}{\dfrac{9}{4}} - \dfrac{(y+2)^2}{5} = 1\)

\((x+3)^2 - \dfrac{1}{5}(y+2)^2 = 1\)

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

\(2x^2 - (y-1)^2 = \dfrac{8}{9}\)

Answer
\(\dfrac{(x^2}{\dfrac{4}{9}} - \dfrac{(y-1)^2}{\dfrac{5}{9}} = 1\)

\(\dfrac{2}{3}(x-5)^2 - \dfrac{5}{3}(y-6)^2 = \dfrac{6}{5}\)

Answer
\(\dfrac{((x-5)^2}{\dfrac{9}{5}} - \dfrac{(y-6)^2}{\dfrac{18}{25}} = 1\)

Subsection 2. Solve a quadratic equation for \(y\)

Subsubsection Example

Example 9.48.

Solve for \(y\text{:}\) \(~\dfrac{y^2}{4}-\dfrac{x^2}{9} = 1\)

Solution

We begin by isolating \(y^2\text{.}\)

\begin{equation*} y^2 = 4 \left(1+\dfrac{x^2}{9}\right) \end{equation*}

Before extracting roots, we simplify the right side of the equation.

\begin{align*} y^2 \amp = 4\left(\dfrac{9+x^2}{9}\right)\\ y^2 \amp = \dfrac{4}{9}(9+x^2) \end{align*}

Finally, we take square roots of both sides, and simplify.

\begin{align*} y \amp = \pm \sqrt{\dfrac{4}{9}(9+x^2)}\\ y \amp = \pm \dfrac{2}{3}\sqrt{9+x^2} \end{align*}

Subsubsection Exercises

Solve each equation for \(y\text{.}\)

\(y^2-4x^2=4\)

Answer
\(y= \pm 2\sqrt{1+x^2}\)

\((x-2)^2-y^2=4\)

Answer
\(y=\pm \sqrt{x^2-4x}\)

\(\dfrac{x^2}{3}+\dfrac{y^2}{12} = 1\)

Answer
\(y=\pm 2\sqrt{3-x^2}\)

\(y^2-6y=x+3~\) (Hint: complete the square in \(y\text{.}\))

Answer
\(y=3 \pm \sqrt{x+12}\)

Subsection 3. Find an asymptote

Subsubsection Example

Example 9.53.

The points \(~(1,3),~(9,3),~(5,10),~\) and \(~(5,-4)~\) are the midpoints of the four sides of a rectangle.

  1. Sketch the rectangle.
  2. Find equations for the diagonals of the rectangle.
Solution
  1. The easiest way find the diagonals of the rectangle is to sketch it first. From the sketch at right, we can see that the center of the rectangle is \((5,3)\text{,}\) and the two upper vertices are \((1,10)\) and \((9,10)\text{.}\)
  2. Both diagonals pass through the center, \((5,3)\text{.}\) The diagonal that passes through \((9,10)\) has slope
    \begin{equation*} m=\dfrac{10-3}{9-5} = \dfrac{7}{4} \end{equation*}
rectangle and diagoals

We use the point-slope formula to find its equation.

\begin{align*} \dfrac{y-3}{x-5} \amp = \dfrac{7}{4} \amp \amp \blert{\text{Cross-multiply.}}\\ 4(y-3) \amp = 7(x-5) \amp \amp \blert{\text{Apply the distributive law.}}\\ 4y-12 \amp = 7x-35 \amp \amp \blert{\text{Add 12 to both sides.}}\\ 4y \amp = 7x-23 \amp \amp \blert{\text{Divide both sides by 4.}}\\ y \amp = \dfrac{7}{4}x - \dfrac{23}{4} \end{align*}

Similarly, you can check that the diagonal that passes through \((1,10)\) has slope \(\dfrac{-7}{4}\text{,}\) and its equation is \(y = \dfrac{-7}{4}x + \dfrac{47}{4}\)

Subsubsection Exercises

The four points given are the midpoints of the four sides of a rectangle.

  1. Sketch the rectangle.
  2. Find equations for the diagonals of the rectangle.

\((-7,-1),~(1,-1),~(-3,-4),~(-3,2)\)

Answer
  1. rectangle and diagoals
  2. \(\displaystyle y=\dfrac{3}{4}x+\dfrac{5}{4};~y=\dfrac{-3}{4}x-\dfrac{13}{4}\)

\((5,-8),~(9,-8),~(7,-15),~(7,-1)\)

Answer
  1. rectangle and diagoals
  2. \(\displaystyle y=\dfrac{7}{2}x-\dfrac{65}{2};~y=\dfrac{-7}{2}x+\dfrac{61}{2}\)