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Form the differential equation of the family of parabolas having vertex at the origin and axis along positive y - axis.
cbse
class12
modelpaper
2012
sec-b
q15
easy
math
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asked
Feb 4, 2013
by
thanvigandhi_1
retagged
Dec 2, 2013
by
sreemathi.v
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Toolbox:
Equation of a parabola having vertex at origin at axis along the positive axis is $x^2 = 4ay $
Step 1:
Given: $ x^2 = 4ay$-------(1)
Differentiating on both sides we get,
$2x = 4ay'$
$x = 2ay'$
$a =\large\frac{ x}{2y'}$
Step 2:
Substituting in equ(1) we get
$x^2 = 4. [\large\frac{x}{2y'}] $$.y$
On simplifying and rearranging we get
$xy' -2y = 0$
This is the required equation
answered
Dec 2, 2013
by
sreemathi.v
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