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Questions  >>  CBSE XII  >>  Math  >>  Differential Equations

Form the differential equation of the family of parabolas with vertex on the x axis with symmetry axis parallel to the axis y positive and the vertex to focus distance equal to P.

I know the equation of this parabola is (x-X0)^2 = 4Py.

Please help me!


1 Answer

The given equation of the parabola is $(x-h)^2=4py$. Since it has two constants we will have to differentiate twice.
Differentiating wrt x we get
I.e. $(x-h)=2py' $
Differentiating once again wrt x we get
$1=2py'' $
Therefore $y''=\large\frac{1}{2p}$
Hence $(x-h)=\large\frac{y'}{y''}$
Substituting this in the first equation we get $ (\large\frac{y'}{y''})^2= (x-h)^2 = 4py = 2p*2y = \large\frac{2y}{y''} $
Therefore on simplifying we get
$(y')^2-2y(y'')=0 $
I don't understand this part: Hence (x-h)=y'/y" Substituting this in the first equation we get (Y'/y")^2=2y/y' Therefore on simplifying we get (Y')^3-2y(y")^2=0 This is the required equation.
hector - pls see answer above.
I've seen. Thank you.
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