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# A satellite is travelling around the earth in an elliptical orbit having the earth heat a focus and of eccentricity $\large\frac{1}{2}.$ The shortest distance that the satellite get to the earth is $400kms.$ Find the longest distance that the satellite gets from the earth.

Toolbox:
• Standard forms of equation of the ellipse with major axes 2a, minor axis 2b $(a >b)$ eccentricity e and $b^2=a^2(1-e^2)$ or $e^2=1-\large\frac{b^2}{a^2}$
• $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 • http://clay6.com/mpaimg/18_2_Toolbox.png • Foci(\pm ae,o), center (0,0),vertices (\pm a,0) • End points of Latus Rectum \; (ae,\pm \large\frac{b^2}{a}) and (-ae,\pm \large\frac{b^2}{a}) • Directrices x=\pm \large\frac{a}{e}. • The major axis is y=0 (x- axis) and the minor axes is x=0 (y- axis) • \large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
• http://clay6.com/mpaimg/18_2_Toolbox1.png
• Foci$(0,\pm ae),$ center $(0,0)$,vertices $(0,\pm a)$
• End points of Latus Rectum $(\pm \large\frac{b^2}{a}$$,ae) and (\pm \large\frac{b^2}{a}$$,-ae)$
• Directrices $y=\pm \large\frac{a}{e}$.
• The major axis is $x=0$ (y- axis) and the minor axes is $y=0$ (x- axis)
• If a point moves so that the sum of its distances from two fixed points is a constant,then the path traced as an ellipse with major axis of length equal to the constant sum and foci at the two fixed points.
Step 1:
Let the foci be at $F(ae,0),F'(-ae,0)$ and the vertices at $A(a,0),A'(-a,0)$.
Let the earth be at $F(ae,0)$.
Step 2:
The shortest distance of the satellite is at $A$
$FA=a-ae=400$
Now $e=\large\frac{1}{2}$
Therefore $a(1-e)=400$
$\large\frac{a}{2}$$=400\Rightarrow a=800$
Step 3:
The maximum distance of the satellite is at $A'$
(i.e) $FA'=a+ae$
$\Rightarrow a(1+e)$.
$\Rightarrow 806\times \large\frac{3}{2}$
$\Rightarrow 1200m$
Therefore the longest distance that the satellite from the earth is 1200 m.
edited Jun 20, 2013