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{\bf Question}

The position vectors of the foci of an ellipse are $\pm{\bf c}$
and the length of the major axis is $2a$.  Show that the equation
of the ellipse is given by $$a^4 - a^2({\bf r}^2 + {\bf c}^2) +
({\bf r} \cdot {\bf c})^2 = 0.$$

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{\bf Answer}

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${}$

$\ds |S_1R| + |S_2R| = 2a$

$\ds |{\bf r - c}|^2 + |{\bf r + c}|^2 = 2a$

$\ds |{\bf r - c}|^2 + |{\bf r+c}|^2 + 2|{\bf r - c}||{\bf r + c}|
= 4a^2$

Now \begin{eqnarray*} |{\bf r - c}|^2 & = & {\bf r \cdot r} -
2{\bf r \cdot c + c \cdot c} \\ |{\bf r + c}|^2 & = & {\bf r \cdot
r} + 2{\bf r \cdot c + c \cdot c} \end{eqnarray*}

${}$

So \begin{eqnarray*} |{\bf r - c}||{\bf r + c}| & = & 2a^2 - (r^2
+ c^2) \\ |{\bf r - c}|^2|{\bf r + c}|^2 & = & (r^2 +c^2 - 2rc)
(r^2 + c^2 + 2rc) \\ & = & (r^2+c^2)^2 - 4(rc)^2 \end{eqnarray*}
So \begin{eqnarray*} (r^2 + c^2)^2 - 4(rc)^2 & = & 4a^4 -
4a^2(r^2+c^2) + (r^2 + c^2)^2 \\ a^4 - a^2(r^2 + c^2) + (rc)^2 & =
& 0 \end{eqnarray*}



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