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The focus of a parabolic mirror is at a distance of $8cm$ from its centre (vertex) ,if the mirror is $25cm$ deep, find the diameter of the mirror.

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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 $V(0,0)$ be the centre of the parabolic mirror whose focus is $8$cm from $V$ (i.e) at $(8,0)$
$VF$ is the axis of the mirror,and it bisects the diameter $AB$ of the mirror at a distance of $25$cm from the focus.
Let $A$ be the point $(25,y)$ which lies on the parabola of which the mirror is a part.
Step 2:
Here $a=8$
Therefore the equation of the parabola is $y^2=4ax$
$\Rightarrow 4\times 8x$
$y^2=32x$
Step 3:
Substituting the coordinates $(25,y)$
$y^2=32\times 25$
$y=\pm \sqrt{32\times 25}$
$\Rightarrow y=\pm 20\sqrt 2$
The points $A$ and $B$ are $(25,20\sqrt 2),(25,-20\sqrt 2)$
The diameter is $AB=40\sqrt 2$cm.
answered Jun 14, 2013 by sreemathi.v
edited Jun 14, 2013 by sreemathi.v
 

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