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Differential Equations
Find nature of curve not passing through origin for which tangent at any point P. on curve is equal to radius vector of point P.
$(a)\;x^2-y^2=c \\ (b)\;y^2+9x^2+6=c \\ (c)\;x^2+3xy^2-y=c \\ (d)\;x^2+3y^2=c $
jeemain
math
class12
ch9
differential-equations
formation-of-differential-equations
q83
difficult
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asked
Feb 7, 2014
by
meena.p
retagged
Oct 26, 2014
by
sharmaaparna1
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1 Answer
Equation of normal at P
$Y-y=-\large\frac{-1}{m}$$(X-x)$
=> $ X+mY=x +my$
$PN=\sqrt {m^2y^2+y^2}$
$OP= PN$
$x^2+y^2=m^2y^2+y^2$
$x^2 =m^2y^2$
$\large\frac{dy}{dx}$$=m =\pm \large\frac{x}{y}$
=> $ \int y dy =\pm \int x dy$
$x^2+y^2=c$ (circle)
or $x^2-y^2=c$
Hence a is the correct answer.
answered
Feb 7, 2014
by
meena.p
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