Graphing hyperbola, locate the foci and asymptotes: 9x^2-16y^2=144?
Good, you got almost all the way by yourself! 9x² - 16y² = 144 Divide the equation by 144 to set the righthand side to 1. 9x²/144 - 16y²/144 = 144/144 x²/16 - y²/9 = 1 The x² term is positive so the hyperbola is horizontal. Standard equation for a horizontal hyperbola:   (x - h)²/a² - (y - k)²/b² = 1 with   center (h, k)   foci (h±c, k) where c² = a² + b²   asymptotes y = ±(b/a)x + k∓(b/a)h x²/4² - y²/3² = 1 center (0, 0) c² = 4² + 3² = 25 c = 5 foci (0±5, 0) = (-5, 0) and (5, 0) asymptotes y = ±(3/4)x + k∓(3/4)0   y = -(3/4)x   y = (3/4)x http://www.flickr.com/photos/dwread/1143...
Hyperbola and Ellipse math help?
Ellipses: the equation for an ellipse is (x-h)^2/a^2+(y-k)^2/b^2=1 (a^2 and b^2 can switch) so for #6 you will have to divide everything by 648 to get the equation to =1. so you end up with: (x^2/36)+(y^2/18)=1 For ellipses, the biggest denominator is a and the other one is b. So for this, 36=a^2 (it is the bigger number) so a=6. b=sqrt18 or 3sqrt2. the ellipse is horizontal because a^2 is under x^2. the center is (0,0). so since the ellipse is horizontal, the vertex and foci will change the x coordinate. the covertex will change the y coordinate. for this ellipse, the vertices are (a,0) and (-a,0). the covertices are (0,b) and (0,-b). so if you plug in a and b you get vertices(6,0) and (-6,0) and covertices (0,3sqrt2) and (0,-3sqrt2). to find the foci, you need to find c. to find c use the equation: c^2=a^2-b^2. if you plug in a and b, you get c^2=28. so c=sqrt28 or 2sqrt7. the foci are (c,0) and (-c,0). so the foci are (2sqrt7,0) and (-2sqrt7,0) then to graph, just graph all of these points if the ellipses were to be vertical, a^2 would be under y^2 and the vertices for center (h,k) would be (h,k+a) and (h,k-a). covertices would be (h+b,k) and (h-b,k). foci would be (h,k+c) and (h,k-c). hyperbolas: here are the basics... horizontal hyperbola equation is ((x-h)^2/a^2)-((y-k)^2/b^2)=1 (it is horizontal b/c x^2 is first. a is always under the first term) center:(h,k) vertices: (h+a,k) and (h-a,k) covertices:: (h,k+b) and (h, k-b) foci: (h+c,k) and (h-c,k) (to find c, use a^2 + b^2= c^2 which is kind of the opposite equation from ellipses) asymptotes: (y-k)=+/-b/a(x-h) (+/- means plus or minus. there will be 2 asymptotes) to graph, draw a box connecting the vertices and covertices then the asymptotes should go thru the center and the corners of the box. to draw the hyperboles, start at the vertices and draw a curved line toward an asymptote but it never touches the asymptote. vertical hyperbola equation is: ((y-k^2)/a^2)-((x-h^2)/b^2)=1 (y^2 is first so its vertical) center:(h,k) vertices:(h, k+/-a) covertices: (h+/-b,k) foci: (h,k+/-c) asymp: (y-k)=+/-a/b(x-h)
9x^2 −16y^2 =144. Additionally, list the vertices, foci, center, length of major axis, and the length of the minor axis of this hyperbola.?
standard form of hyperbola (horizontal axis) (x-h)^2 / a^2 - (y-k)^2 / b^2 = 1 asymptote slope m = ± b/a standard form of hyperbola (vertical axis) (y-k)^2 / a^2 - (x-h)^2 / b^2 = 1 asymptote slope m = ± a/b (h,k) => center a: distance between center and vertices c^2 = a^2 + b^2 c: distance between center and focus .......................................... c = √(4^2 + 3^2) = ±5 the foci (±5, 0) http://www.wolframalpha.com/input/?i=plo...