Solutions for The Hyperbola
Solutions to Try Its
1. Vertices: [latex]\left(\pm 3,0\right)[/latex]; Foci: [latex]\left(\pm \sqrt{34},0\right)[/latex] 2. [latex]\frac{{y}^{2}}{4}-\frac{{x}^{2}}{16}=1[/latex] 3. [latex]\frac{{\left(y - 3\right)}^{2}}{25}+\frac{{\left(x - 1\right)}^{2}}{144}=1[/latex] 4. vertices: [latex]\left(\pm 12,0\right)[/latex]; co-vertices: [latex]\left(0,\pm 9\right)[/latex]; foci: [latex]\left(\pm 15,0\right)[/latex]; asymptotes: [latex]y=\pm \frac{3}{4}x[/latex];
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Solutions to Odd-Numbered Exercises
1. A hyperbola is the set of points in a plane the difference of whose distances from two fixed points (foci) is a positive constant. 3. The foci must lie on the transverse axis and be in the interior of the hyperbola. 5. The center must be the midpoint of the line segment joining the foci. 7. yes [latex]\frac{{x}^{2}}{{6}^{2}}-\frac{{y}^{2}}{{3}^{2}}=1[/latex] 9. yes [latex]\frac{{x}^{2}}{{4}^{2}}-\frac{{y}^{2}}{{5}^{2}}=1[/latex] 11. [latex]\frac{{x}^{2}}{{5}^{2}}-\frac{{y}^{2}}{{6}^{2}}=1[/latex]; vertices: [latex]\left(5,0\right),\left(-5,0\right)[/latex]; foci: [latex]\left(\sqrt{61},0\right),\left(-\sqrt{61},0\right)[/latex]; asymptotes: [latex]y=\frac{6}{5}x,y=-\frac{6}{5}x[/latex] 13. [latex]\frac{{y}^{2}}{{2}^{2}}-\frac{{x}^{2}}{{9}^{2}}=1[/latex]; vertices: [latex]\left(0,2\right),\left(0,-2\right)[/latex]; foci: [latex]\left(0,\sqrt{85}\right),\left(0,-\sqrt{85}\right)[/latex]; asymptotes: [latex]y=\frac{2}{9}x,y=-\frac{2}{9}x[/latex] 15. [latex]\frac{{\left(x - 1\right)}^{2}}{{3}^{2}}-\frac{{\left(y - 2\right)}^{2}}{{4}^{2}}=1[/latex]; vertices: [latex]\left(4,2\right),\left(-2,2\right)[/latex]; foci: [latex]\left(6,2\right),\left(-4,2\right)[/latex]; asymptotes: [latex]y=\frac{4}{3}\left(x - 1\right)+2,y=-\frac{4}{3}\left(x - 1\right)+2[/latex] 17. [latex]\frac{{\left(x - 2\right)}^{2}}{{7}^{2}}-\frac{{\left(y+7\right)}^{2}}{{7}^{2}}=1[/latex]; vertices: [latex]\left(9,-7\right),\left(-5,-7\right)[/latex]; foci: [latex]\left(2+7\sqrt{2},-7\right),\left(2 - 7\sqrt{2},-7\right)[/latex]; asymptotes: [latex]y=x - 9,y=-x - 5[/latex] 19. [latex]\frac{{\left(x+3\right)}^{2}}{{3}^{2}}-\frac{{\left(y - 3\right)}^{2}}{{3}^{2}}=1[/latex]; vertices: [latex]\left(0,3\right),\left(-6,3\right)[/latex]; foci: [latex]\left(-3+3\sqrt{2},1\right),\left(-3 - 3\sqrt{2},1\right)[/latex]; asymptotes: [latex]y=x+6,y=-x[/latex] 21. [latex]\frac{{\left(y - 4\right)}^{2}}{{2}^{2}}-\frac{{\left(x - 3\right)}^{2}}{{4}^{2}}=1[/latex]; vertices: [latex]\left(3,6\right),\left(3,2\right)[/latex]; foci: [latex]\left(3,4+2\sqrt{5}\right),\left(3,4 - 2\sqrt{5}\right)[/latex]; asymptotes: [latex]y=\frac{1}{2}\left(x - 3\right)+4,y=-\frac{1}{2}\left(x - 3\right)+4[/latex] 23. [latex]\frac{{\left(y+5\right)}^{2}}{{7}^{2}}-\frac{{\left(x+1\right)}^{2}}{{70}^{2}}=1[/latex]; vertices: [latex]\left(-1,2\right),\left(-1,-12\right)[/latex]; foci: [latex]\left(-1,-5+7\sqrt{101}\right),\left(-1,-5 - 7\sqrt{101}\right)[/latex]; asymptotes: [latex]y=\frac{1}{10}\left(x+1\right)-5,y=-\frac{1}{10}\left(x+1\right)-5[/latex] 25. [latex]\frac{{\left(x+3\right)}^{2}}{{5}^{2}}-\frac{{\left(y - 4\right)}^{2}}{{2}^{2}}=1[/latex]; vertices: [latex]\left(2,4\right),\left(-8,4\right)[/latex]; foci: [latex]\left(-3+\sqrt{29},4\right),\left(-3-\sqrt{29},4\right)[/latex]; asymptotes: [latex]y=\frac{2}{5}\left(x+3\right)+4,y=-\frac{2}{5}\left(x+3\right)+4[/latex] 27. [latex]y=\frac{2}{5}\left(x - 3\right)-4,y=-\frac{2}{5}\left(x - 3\right)-4[/latex] 29. [latex]y=\frac{3}{4}\left(x - 1\right)+1,y=-\frac{3}{4}\left(x - 1\right)+1[/latex] 31.
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- Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.