Math II: Circle, ellipse, parabola/hyperbola eq.

<p>Barrons covers equations of a circle, ellipse, parabola, and hyperbola. The section includes completing the square, finding the center, major/minor axis, and foci.</p>

<p>Edit: Barrons also covers transverse axis, conjugate axis, and eccentricity. Sheesh this book is super in depth :(</p>

<p>Do we need to know this stuff for Math II?</p>

<p>I have never seen a question ask about eccentricity, conjugate axis or transverse axis. However, I have seen questions on completing the square finding the center and on major/minor axis. Not sure about foci though. I think barrons has more information than you need to know. Can anyone confirm this?</p>

<p>You should probably know how to derive the equation of a hyperbola/ellipse/parabola/circle from its graph, because once you know that you can pretty much do everything else that relates to them too.</p>

<p>how do you derive the equation of a hyperbola and ellipse? We never learned that in class.</p>

<p>I'm not so sure on Hyperbolas, but for ellipses, you find the center, the length of the major and minor axes (which is 2a) and then just put them in the standard equation. If it asks for the foci, use the formula c^2=a^2+b^2, if the ellipse is longer horizontally, the foci are (c,0) (-c,0) (if centered at the origin). If it isn't centered at the origin, just move up/down the c value. Could someone please explain Hyperbolas?</p>

<p>sorry if this is confusing!</p>

<p>the equation for a hyperbola is either
(x-h)^2/a^2 - (y-k)^2/b = 1<br>
(x-h)^2/b - (y-k)^2/a = 1
whichever grouping has "a^2" under it is what determines whether the axis is horizontal or vertical. in the first one, the axis is horizontal because a is under (x-h)^2 so the x makes it horizontal. in the second one "a^2" is under (y-k)^2 so the axis is vertical. </p>

<p>the vertices are both "a" units from the center. remember that in the equation it's a^2, so you have to take the square root.
the foci are "c" unites from the center, and since that's not given in the equation you have to find it with the equation: c^2 = a^2 + b^2 and use the "a^2" and "b^2" values from the equation.</p>

<p>an ellipse is pretty similar.
the equation is either
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
(x-h)^2/b^2 + (y-k)^2/a^2
it's practically the same thing as a hyperbola, except instead of subtracting you are adding.
to find the foci "c" , you use the equation c^2 = a^2 - b^2 which is again, the same as the hyperbola equation except this time subtracting.
the length of the major axis = 2a
the length of the minor axis = 2b
so again, if y has a^2 under it, the ellipse is going to be vertical
and if the x has a^2 under it, the ellipse will be horizontal</p>

<p>i hope this helps? sorry if i explained it really badly, but i tried.</p>

<p>Oh never mind it's not so hard... it's almost the same thing as an Ellipse, only the center's x/y coordinate should form a vertical/horizontal line (asymptote) depending on whether the hyperbola is vertical or horizontal. But could someone explain what the a^2 and b^2 as denominators of the equation mean?</p>

<p>So in a Hyperbola, if "2a" is the distance between the two vertices, then what is "b"? And that explanation was good :D</p>

<p>the denominators a^2 and b^2 don't really mean anything specific that i know of. you just use them to draw/solve for other things. like the vertices will always be "a" units from the center, and the foci are always "c" unites from the center, so you use the equation c^2 = a^2 + b^2</p>

<p>let's pretend that a hyperbola opens sideways (axis = horizontal)
the "b" value would be how many points you go up/down. once you have these two points, you can draw a 'box' connecting the vertices to the 'b' points (i don't know what they're called). the diagonals of this box are the asymptotes. that's the only reason besides for use in the formula that i can think of for "b".
i don't think you're going to be able to do this though.</p>

<p>Okay I get what you mean, cause in Ellipses a and b represent half of the major and minor axes, but in Hyperbolas you only have 2a as the distance between vertices. But I don't think matching an equation to a hyperbola graph is common, and even if it was I think it would just be simple (center, distance between vertices) :P Hopefully.... haha</p>

<p>If the hyperbola opens vertically, does the b value become the distance from center to the vertices and a values are now the vertical points to draw the box? Or does the b value become the horizontal distance to draw the box and the a value the distance from center to vertices?</p>

<p>Edit: Nevermind, I figured it out. Since the (x-h)^2 and (y-h)^2 switch positions in the vertical hyperbola, the a value is now under the y value and the b value is under the x, so it kinda helps you memorize that the a value tells you the distance from center to vertices (vertically) and b value tells you the distance to draw the square.</p>