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Show that the locus of a point which moves so that the difference of its distance from the points $(5 , 0 ) $ and $(-5 , 0 )$is $8 $ is $9x^{2}-16y^{2}=144$

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  • If a point moves so that the difference between its distances from the fixed points is a constant,then the path traced by the point is a hyperbola with the length of the transverse axis equal to the constant difference and the two points as foci.
Step 1:
The point $P(x,y)$ moves such that the difference of its distances between the points $F(5,0)$ and $F'(-5,0)$ is $8$.
Therefore it traces a hyperbola with length of the transverse axis $2a=8\Rightarrow a=4$
The foci $FF'$ centre of the hyperbola is the midpoint of $FF'$ (i.e)$(0,0)$
Step 2:
The transverse axis contains the points $F$ and $F'$.Therefore it is the $x$-axis.
$FF'=2ae$
Therefore $2ae=10$
$\Rightarrow ae=5$
$e=\large\frac{5}{a}=\frac{5}{4}$
Step 3:
The semi conjugate axes ' b' is given by
$b=a\sqrt{e^2-1}=4\sqrt{\large\frac{25}{16}-\normalsize 1}$
Step 4:
The equation of the hyperbola is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$
$\large\frac{x^2}{16}-\frac{y^2}{9}$$=1$
(i.e)$9x^2-16y^2=144$
answered Jun 19, 2013 by sreemathi.v
 

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