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If a parabolic reflector is $20cm$ in diameter and $5cm$ deep, find the distance of the focus from the centre of the reflector.

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Toolbox:
  • Standard parabolas :
  • (i)$y^2=4ax$
  • In this case the axis of symmetry is $x$-axis.
  • The focus is $F(a,0)$ and the equation of the directrix is $x=-a$
  • (ii)$y^2=-4ax$
  • In this case the focus is $F(-a,0)$ and the equation of directrix is $x=a$
  • (iii)$x^2=4ay$
  • In this case the axis of symmetry is $y$-axis.
  • The focus is $F(0,a)$ and the equation of directrix is $y=-a$
  • (iv)$x^2=-4ay$
  • In this case the focus is $F(0,-a)$ and the equation of directrix is $y=a$
Step 1:
Let $AB$ be the diameter of the parabolic reflector and $V$ its centre.
The perpendicular bisector of $AB$ passes through the centre $V$
Let $V$ be the origin.The perpendicular bisector of $AB$ is the axis of the parabola of which the mirror is a part.
The coordinates of $A$ and $B$ are $(5,10),(5,-10)$
This parabola opens out to the right and the points $AB$ be on it.
Step 2:
Let its equation be $y^2=4ax$
Substituting $(5,10)$ in the equation,we have
$100=20a$
$a=5$
Therefore the equation of the parabola is $y^2=20x$
The focus is at $(5,0)$.
Therefore the distance of the focus from the centre is $5$cm.
answered Jun 14, 2013 by sreemathi.v
edited Jun 14, 2013 by sreemathi.v
 

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