Conic problem

Captain_P · P versus NP Offline Joined: 27 Jan 2009 Posts: 22

Learning conics for the first time...

Find the equation of the ellipse containing points $(6, - 1)$ , $( - 4, - 5)$ , $(6, - 5)$ , and $( - 12, - 3)$ .

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**Posted:** Sat Nov 07, 2009 9:56 pm

Dr Sonnhard Graubner

hello, we have the equation of the given ellipse as
$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$ , inserting the cordinates of all given points in this equation we get an equation system, solving this we get $a=13,b=\frac{13}{6},x_0=1,y_0=-3$ .
Sonnhard.

**Posted:** Sun Nov 08, 2009 7:16 am

TZF

Right -- assuming that the axes are parallel to the x and y axes, and assuming your observations to be true (which is reasonable if you graph all four points and see where the ellipse should be), you then:

1. Suppose that $(x_0, y_0)=(1,-3)$ , the center.
2. Suppose that semi-axis in the x direction has length equal to the distance from center $(1,-3)$ to vertex $(-12,-3)$ , that is $a=13$ .

Then, plug in any one of the points to solve for $b$ .

The thing is, your first hypothesis is far from true. Consider a right triangle with right-angle vertex at an endpoint of the minor axis, and symmetric over the minor axis (ie, a 45-45-90 triangle). So long as the ellipse is not also a circle, the resulting right triangle contains the center inside of it. There are many, many other such right triangles.

**Posted:** Sun Nov 08, 2009 12:24 pm

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