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Conics:
Hyperbolas: Finding Information Sections: Introduction, Finding information from the equation, Finding the equation from information
Since the y part of the equation is added, then the center, foci, and vertices will be above and below the center (on a line paralleling the yaxis), rather than side by side. Looking at the denominators, I see that a^{2} = 25 and b^{2} = 144, so a = 5 and b = 12. The equation c^{2} – a^{2} = b^{2} tells me that c^{2} = 144 + 25 = 169, so c = 13, and the eccentricity is e = 13/5. Since x^{2} = (x – 0)^{2} and y^{2} = (y – 0)^{2}, then the center is at (h, k) = (0, 0). The vertices and foci are above and below the center, so the foci are at (0, –13) and (0, 13), and the vertices are at (0, 5) and (0, –5).
And the rest of my answer is: center (0,
0), vertices
(0,
–5) and (0,
5), foci
(0,
–13) and
(0,
13),
Since the x part is added, then a^{2} = 16 and b^{2} = 9, so a = 4 and b = 3. Also, this hyperbola's foci and vertices are to the left and right of the center, on a horizontal line paralleling the xaxis. From the equation, clearly the center is at (h, k) = (–3, 2). Since the vertices are a = 4 units to either side, then they are at (–7, 2) and at (1, 2). The equation c^{2} – a^{2} = b^{2} gives me c^{2} = 9 + 16 = 25, so c = 5, and the foci, being 5 units to either side of the center, must be at (–8, 2) and (2, 2). Copyright © Elizabeth Stapel 20102011 All Rights Reserved Since the a^{2} went with the x part of the equation, then a is in the denominator of the slopes of the asymptotes, giving me m = ± 3/4. Keeping in mind that the asymptotes go through the center of the hyperbola, the asymptes are then given by the straightline equations y – 2 = ± (3/4)(x + 3). center (–3,
2), vertices
(–7,
2) and (1,
2), foci
(–8,
2) and (2,
2),
To find the information I need, I'll first have to convert this equation to "conics" form by completing the square. 4x^{2} + 40x
– 5y^{2} – 30y = 45
Then the center is at (h, k) = (–5, –3). Since the x part of the equation is added, then the center, foci, and vertices lie on a horizontal line paralleling the xaxis; a^{2} = 25 and b^{2} = 20, so a = 5 and b = 2sqrt[5]. The equation a^{2} + b^{2} = c^{2} gives me c^{2} = 25 + 20 = 45, so c = sqrt[45] = 3sqrt[5]. The slopes of the two asymptotes will be m = ± (2/5)sqrt[5]. Then my complete answer is: center (–5, –3), vertices (–10, –3) and (0, –3), foci and , and asymptotes If I had needed to graph this hyperbola, I'd have used a decimal approximation of ± 0.89442719... for the slope, but would have rounded the value to something reasonable like m = ± 0.9. << Previous Top  1  2  3  Return to Index Next >>



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