Question 311310
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General equation of a horizontally-oriented ellipse with center at *[tex \Large \left(h,k\right)], semi-major axis *[tex \Large a] and semi-minor axis *[tex \Large b] where *[tex \Large a\ >\ b]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\,-\,h\right)^2}{a^2}\ +\ \frac{\left(y\,-\,k\right)^2}{b^2}\ =\ 1]


So if the major axis measures 20, then *[tex \Large a\ =\ 10], if the minor axis measures 8, then *[tex \Large b\ =\ 4], and if it is centered at the origin, *[tex \Large h\ =\ 0] and *[tex \Large k\ =\ 0], therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x,y)\ =\ \frac{\left(x\,-\,0\right)^2}{10^2}\ +\ \frac{\left(y\,-\,0\right)^2}{4^2}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x,y)\ =\ \frac{x^2}{100}\ +\ \frac{y^2}{16}\ =\ 1]


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A circle is just a special case of an ellipse with the following three characteristics:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ]The measure of the major axis is equal to the measure of the minor axis.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ]The two foci of the ellipse are coincident, and that point is called the center of the circle.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ]The eccentricity, which is the ratio of the distance between the foci and the measure of the major axis, is zero.


All of that makes perfectly good sense algebraically.  Consider the general equation of an ellipse:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\,-\,h\right)^2}{a^2}\ +\ \frac{\left(y\,-\,k\right)^2}{b^2}\ =\ 1]


Now consider this ellipse where the measures of the major and minor axes are equal, which is to say:  *[tex \Large a\ =\ b].  In this case, we can re-write the general equation thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\,-\,h\right)^2}{a^2}\ +\ \frac{\left(y\,-\,k\right)^2}{a^2}\ =\ 1]


And then multiply both sides by *[tex \Large a^2]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\,-\,h\right)^2\ +\ \left(y\,-\,k\right)^2\ =\ a^2]


If you allow *[tex \Large a] to represent the radius of the circle, since *[tex \Large a] is one-half of either of the identical axes of this special ellipse, and that makes either of the axes be the diameter of the circle hence anything that is half of the diameter is by definition a radius, then you can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\,-\,h\right)^2\ +\ \left(y\,-\,k\right)^2\ =\ r^2]



Which is exactly the general equation of a circle centered at *[tex \Large \left(h,k\right)] with radius *[tex \Large r].


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\,+\,3\right)^2}{9} + \frac{y^2}{25}\ =\ 1 ]


Re-write to fit the pattern of the general form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\,+\,3\right)^2}{9} + \frac{\left(y\,-\,0\right)^2}{25}\ =\ 1 ]


Note that for this ellipse *[tex \Large a\ <\ b].  That means that the ellipse is oriented vertically.  So, by examination:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ -3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ]Center: *[tex \Large \left(-3,0\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ]Minor Axis: 6


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ]Major Axis: 10


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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