284. Ellipse (definition).
by Virgil Nicula, Jun 12, 2011, 1:49 PM
The foci of the ellipse are two special points 
and 
on the ellipse's major axis and are equidistant from the center point 
.
The sum of the distances from any point
on the ellipse to those two foci is constant and equal to the major diameter, i.e. 
.
Therefore,
. I"ll prove that 
. Indeed,
![$\boxed{\ \begin{array}{c}
\\
\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a\\\\
\left[\sqrt{(x+c)^2 + y^2}+\sqrt{(x-c)^2 + y^2}\right]\left[\sqrt{(x+c)^2 + y^2}-\sqrt{(x-c)^2+y^2}\right] =2a\left[\sqrt{(x+c)^2 + y^2}-\sqrt{(x-c)^2 + y^2}\right]\\\\
(x+c)^2-(x-c)^2=2a\left[\sqrt{(x+c)^2 + y^2}-\sqrt{(x-c)^2 + y^2}\right]\\\\
\left\{\begin{array}{c}
\sqrt{(x+c)^2 + y^2}-\sqrt{(x-c)^2 + y^2}=\frac {2cx}{a}\\\\
\sqrt{(x +c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a\end{array}\right\|\ \bigoplus\\\\
\sqrt{(x+c)^2 + y^2}=\frac {cx}{a}+a\\\\
(x+c)^2 + y^2=\left(\frac {cx}{a}\right)^2+2cx+a^2\\\\
x^2+c^2+y^2=\frac {c^2x^2}{a^2}+a^2\\\\
\left(a^2-c^2\right)x^2+a^2y^2=a^2\left(a^2-c^2\right)\\\\
\boxed{\ a^2-c^2=b^2\ }\implies b^2 x^2+a^2 y^2=a^2b^2\\\\
\boxed{\frac {x^2}{a^2}+\frac {y^2}{b^2}=1}\\\\
\end{array}\ }$](//latex.artofproblemsolving.com/8/a/c/8acd65568aa29df557654df75ad3ae311f56ecf9.png)
.
PP1. An ellipse has focus
and 
and is tangent to the line 
. Find the equation of this ellipse.
Proof. Observe that the center of this ellipse is
. Denote 
and 
. Therefore, 
and the equation of the required ellipse is
. Denote 
. Thus, the line 
is tangent to the ellipse with equation 
the equation 
has 
the equation 
has the discriminant 
and in this case for 
have 
and 
. In conclusion,
the equation of the required ellipse is
and the line with the equation 
is tangent to this ellipse in the point 
.
and 
on the ellipse's major axis and are equidistant from the center point 
.The sum of the distances from any point
on the ellipse to those two foci is constant and equal to the major diameter, i.e. 
.Therefore,
. I"ll prove that 
. Indeed,![$\boxed{\ \begin{array}{c}
\\
\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a\\\\
\left[\sqrt{(x+c)^2 + y^2}+\sqrt{(x-c)^2 + y^2}\right]\left[\sqrt{(x+c)^2 + y^2}-\sqrt{(x-c)^2+y^2}\right] =2a\left[\sqrt{(x+c)^2 + y^2}-\sqrt{(x-c)^2 + y^2}\right]\\\\
(x+c)^2-(x-c)^2=2a\left[\sqrt{(x+c)^2 + y^2}-\sqrt{(x-c)^2 + y^2}\right]\\\\
\left\{\begin{array}{c}
\sqrt{(x+c)^2 + y^2}-\sqrt{(x-c)^2 + y^2}=\frac {2cx}{a}\\\\
\sqrt{(x +c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a\end{array}\right\|\ \bigoplus\\\\
\sqrt{(x+c)^2 + y^2}=\frac {cx}{a}+a\\\\
(x+c)^2 + y^2=\left(\frac {cx}{a}\right)^2+2cx+a^2\\\\
x^2+c^2+y^2=\frac {c^2x^2}{a^2}+a^2\\\\
\left(a^2-c^2\right)x^2+a^2y^2=a^2\left(a^2-c^2\right)\\\\
\boxed{\ a^2-c^2=b^2\ }\implies b^2 x^2+a^2 y^2=a^2b^2\\\\
\boxed{\frac {x^2}{a^2}+\frac {y^2}{b^2}=1}\\\\
\end{array}\ }$](http://latex.artofproblemsolving.com/8/a/c/8acd65568aa29df557654df75ad3ae311f56ecf9.png)
.PP1. An ellipse has focus
and 
and is tangent to the line 
. Find the equation of this ellipse.Proof. Observe that the center of this ellipse is
. Denote 
and 
. Therefore, 
and the equation of the required ellipse is
. Denote 
. Thus, the line 
is tangent to the ellipse with equation 
the equation 
has 
the equation 
has the discriminant 
and in this case for 
have 
and 
. In conclusion,the equation of the required ellipse is
and the line with the equation 
is tangent to this ellipse in the point 
.This post has been edited 16 times. Last edited by Virgil Nicula, Nov 21, 2015, 5:45 PM
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