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Conics: Hyperbolas (page 3 of 3) Sections: Introduction, Finding information from the equation, Finding the equation from information
The center, focus, and vertex all lie on the horizontal line y = 3 (that is, they're side by side on a line paralleling the xaxis), so the branches must be side by side, and the x part of the equation must be added. The a^{2} will go with the x part of the equation, and the y part will be subtracted. The vertex is 2 units from the center, so a = 2; the focus is 3 units from the center, so c = 3. Then a^{2} + b^{2} = c^{2} gives me b^{2} = 9 – 4 = 5. I don't need to bother with the value of b itself, since they only asked me for the equation, which is:
The vertex and the center are both on the vertical line x = 0 (that is, on the yaxis), so the hyperbola's branches are above and below each other, not side by side. Then the y part of the equation will be added, and will get the a^{2} as its denominator. Also, the slopes of the two asymptotes will be of the form m = ± a/b. The vertex they gave me is 5
units above the center, so a
= 5 and a^{2}
= 25. The slope of the asymptotes
(ignoring the "plusminus" part) is a/b
= 5/3 = 5/b, so b
= 3 and b^{2}
= 9. And this is all I need in order
to find my equation:
The foci are side by side, so this hyperbola's branches are side by side, and the center, foci, and vertices lie on a line paralleling the xaxis. So the y part of the equation will be subtracted and the a^{2} will go with the x part of the equation. The center is midway between the foci, so the center must be at (h, k) = (–1, 0). The foci are 5 units to either side of the center, so c = 5 and c^{2} = 25. Copyright © Elizabeth Stapel 20102011 All Rights Reserved The center lies on the xaxis, so the two xintercepts must then also be the hyperbola's vertices. Since the intercepts are 4 units to either side of the center, then a = 4 and a^{2} = 16. Then a^{2} + b^{2} = c^{2} tells me that b^{2} = 25 – 16 = 9, and my equation is:
The vertices are above and below each other, so the center, foci, and vertices lie on a vertical line paralleling the yaxis. Then the a^{2} will go with the y part of the hyperbola equation, and the x part will be subtracted. The center is midway between the two vertices, so (h, k) = (–2, 7). The vertices are 8 units above and below the center, so a = 8 and a^{2} = 64. The eccentricity is e = c/a = 17/8 = c/8, so c = 17 and c^{2} = 289. The equation a^{2} + b^{2} = c^{2} tells me that b^{2} = 289 – 64 = 225. Then my equation is: << Previous Top  1  2  3  Return to Index



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