Graphing a Hyperbola Centered at (0, 0) OBJECTIVE Sketch the graph of a hyperbola in the form (i) x²/a² - y²/b² = 1 or (ii) y²/a² - x²/b² = 1. Step 1 Write the equation in standard form. Determine transverse axis and the orientation of the hyperbola. Step 2 Locate vertices and the endpoints of

Question

Graphing a Hyperbola Centered at (0, 0)

OBJECTIVE

Sketch the graph of a hyperbola in the form

(i) [latex]\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1[/latex] or (ii) [latex] \frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1[/latex].

Step 1 Write the equation in standard form. Determine transverse axis and the orientation of the hyperbola.

Step 2 Locate vertices and the endpoints of the conjugate axis.

Step 3 Lightly sketch the fundamental rectangle by drawing dashed lines parallel to the coordinate axes through the points in Step 2.

Step 4 Sketch the asymptotes. Extend the diagonals of the fundamental rectangle. These are the asymptotes.

Step 5 Sketch the graph. Draw both branches of the hyperbola through the vertices, approaching the asymptotes. See Figures 25 and 26.The foci are located on the transverse axis, c units from the center, where c² = a² +b².

Sketch the graph of

a. 16x² - 9y² = 144. b. 25y² - 4x² = 100.

Accepted Answer

[latex]\begin{aligned}16 x^{2}-9 y^{2} &=144 \\frac{16 x^{2}}{144}-\frac{9 y^{2}}{144} &=\frac{144}{144} \\frac{x^{2}}{9}-\frac{y^{2}}{16} &=1 \end{aligned}[/latex]

Minus sign precedes y²-term, and transverse axis is along the x-axis; hyperbola opens left and right.

Because [latex] a^{2}=9, a=3 .[/latex] Also, [latex] b^{2}=16 ; ext { so } b=4 [/latex]. The vertices are (±3, 0), and the endpoints of the conjugate axis are (0, ±4).

Draw dashed lines x = 3, x = -3, y = 4,and y = -4 to form the fundamental rectangle with vertices (3, 4), (-3, 4), (-3, -4), and (3, -4).

[latex]\begin{aligned}25 y^{2}-4 x^{2} &=100 \\frac{25 y^{2}}{100}-\frac{4 x^{2}}{100} &=\frac{100}{100} \\frac{y^{2}}{4}-\frac{x^{2}}{25} &=1\end{aligned}[/latex]

Minus sign precedes x²-term, and transverse axis is along the y-axis; hyperbola opens up and down.

Because [latex] a^{2}=4, a=2.[/latex] Also, [latex] b^{2}=25 ; ext { so } b=5.[/latex] The vertices are (0, ±2) and the endpoints of the conjugate axis are (±5, 0).

Draw dashed lines y = 2, y = -2, x = 5, and x = -5 to form the fundamental rectangle with vertices (5, 2), (-5, 2), (-5, -2), and (5, -2).

Question 11.4.5: Graphing a Hyperbola Centered at (0, 0) OBJECTIVE Sketch the...

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