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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of parabola whose vertex is (4,2) and focus is (6,2)?

$\begin{array}{1 1}(a)\;(y-3)^2=8(k-4)\\(b)\;(y-2)^2=8(k-4)\\(c)\;(y-2)^2=4(k-4)\\(d)\;(y-2)^2=6(k-4)\end{array}$

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1 Answer

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Let vertex P(4,2) and focus Q is (6,2)
Slope of PQ=0
Hence axis of parabola parallel to x-axis
The equation is of the form
$(y-k)^2=4a(x-h)$
(h,k) is vertex
$(y-2)^2=4a(x-4)$
$a=\sqrt{(6-4)^2+(2-2)^2}$
$a=2$
Hence equation of parabola is
$(y-2)^2=8(k-4)$
Hence (b) is the correct answer.
answered Feb 6, 2014 by sreemathi.v
 

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