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Find the locus of the point which moves so that the sum of its distances from $(3 , 0 )$ and $(-3 , 0 ) $ is $ 9.$

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  • If a point moves so that the sum of its distances from the 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 $P(x,y)$ be the point which moves such that sum of its distances from $F(3,0)$ and $F'(-3,0)$ is $9$
The curve traced by $P$ is an ellipse length of whose major axes $(2a=9)$ and whose central is at the midpoint of $FF'$ at $(0,0)$,where $F,F'$ are the foci.
Step 2:
The major axis is the $x$-axis since the foci lie on it.
Now $CF=ae=3$
$a=\large\frac{9}{2}$
Therefore $e=\large\frac{3}{a}=\frac{3\times 2}{9}=\frac{2}{3}$
Step 3:
$b=a\sqrt{1-e^2}$
$\Rightarrow \large\frac{9}{2}\sqrt{1-\large\frac{4}{9}}=\frac{3\sqrt 5}{2}$
$b^2=\large\frac{45}{4}$
The equation is $\large\frac{x^2}{81/4}+\frac{y^2}{45/4}$$=1$
answered Jun 17, 2013 by sreemathi.v
 

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