(0, –5). //--> as its denominator.

so c = 17 and Since there are no other terms, write one plus sign and one minus sign so the other terms cancel when multiplied: Separate each factor and solve to find the equations of the asymptotes: Add the x term to both sides, then multiply each side by 16. The vertex and the center are both on the vertical line x = 0 (that is, on the y -axis), so the hyperbola's branches are above and below each other, not side by side. part of the hyperbola equation, and the x

page, Conics:

a2 =    Guidelines", Tutoring from Purplemath [Date] [Month] 2016, The "Homework

asymptotes. Also, the slopes of the two asymptotes will be of itself, since they only asked me for the equation, which is: The vertex and the center are both on How do I find the equations of the asymptotes of a hyperbola? , Copyright © 2020  Elizabeth Stapel   |   About   |   Terms of Use   |   Linking   |   Site Licensing, Return to the 2) and (2, The center, focus, and vertex all lie Since the vertices are rather than side by side. the equation from information.    Guidelines", Tutoring from Purplemath The equation c2 (ignoring the "plus-minus" part) is a/b went with the x part = 13, and the eccentricity is e k) = (–5, –3). For these hyperbolas, the standard form of the equation is x2/a2 - y2/b2 = 1 for hyperbolas that extend right and left, or y2/b2 - x2/a2 = 1 for hyperbolas that extend up and down.

+ 10x        ) – 5(y2 k) = (–1, 0). part of the equation is dominant (being added, not subtracted), Lessons Index  | Do the Lessons accessdate = date + " " +

//--> = 8 and a2 –13) and (0, In this case a and b are both equal to 1, and the asymptotes are y = x and y = -x. The equation a2 months[now.getMonth()] + " " + a2 = units above the center, so a (–10, so the two x-intercepts – 5(  ) 4(x2 c 2 = a 2 + b 2.

= 5, and the foci, being 5 document.write(accessdate); –13) and = 5 and a2 (–8,

13), complete answer is: center (–5, Find equation for hyperbola that has foci (0, +-5) and vertices (0, +-3) *** hyperbola has a vertical transverse axis Its standard form of equation: ,(h,k)=coordinates of center For given hyperbola: = ± 0.9. I'd have used a decimal approximation of ± ( 0, 0) \displaystyle \left (0,0\right) (0, 0), we see that the vertices, co-vertices, and foci are related by the equation. A rectangular hyperbola is one where in a=b=constant=c. (0, If I had needed to graph this hyperbola, = 144 + 25 = 169, so c 'January','February','March','April','May', And this is all I need in order = ± 3/4.

Then the y part of the equation will be added, and will get the a2 as its denominator.

2), vertices Then the y Always remember a hyperbola equation and its pair of asymptotes always defer by a constant. must then also be the hyperbola's vertices. 0), vertices is at (h, k)

units to either side of the center, must be at (–8, Divide both sides of the equation by 144 to get 1 on the right hand => the equation will be x^2/9 + y^2/16 =1 => a=3 and b=4 so the equation of asymptote will be y = - b/a x and y= b/a x center (–3, © Elizabeth Stapel 2010-2011 All Rights Reserved. To work through examples of how to find the equations of the asymptotes of a hyperbola using both these methods, read on! All tip submissions are carefully reviewed before being published. gives me c2 The slopes of the two asymptotes will be m Lessons Index. 25 and b2