logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

A kho-kho player in a practice session while running realises that the sum of the distances from the two kho-kho poles from him is always $8m$ .Find the equation of the path traced by him if the distance between the poles is $6m$.

Can you answer this question?
 
 

1 Answer

0 votes
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}$$=1$
  • http://clay6.com/mpaimg/17_2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $\; (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/17_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:
Since the sum of his distances from the two poles is a constant (8 m) the player is tracing an ellipse with the two poles as foci.
The sum of the distances =$FP+F'P=2a=8$
Therefore $a=4m$
Also $FF'$=distance between the poles =$2ae$
$2ae=6\Rightarrow 2\times 4e=6$
$\Rightarrow e=\large\frac{3}{4}$
Step 2:
The path traced is an ellipse with $a=4,e=\large\frac{3}{4}$
Therefore $b=a\sqrt{1-e^2}$
$\Rightarrow 4\sqrt{1-\large\frac{9}{16}}=\sqrt 7$
The equation of the ellipse is $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
$\Rightarrow \large\frac{x^2}{16}+\frac{y^2}{7}$$=1$
answered Jun 18, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...