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The tangent to a curve intersects x axis at $B$ and $y$ at $A$. The normal to same curve intersects $x$ axis at $C$ and $Y$ axis at $D$. Given $\large\frac{AO}{OB}= \frac{OC}{CD}$ curve passes through $(2,3)$. Find the curve

$(a)\;y^2=-x^3+9x^2+t \\(b)\;y^2=-x^4+\frac{25}{4}x^2 \\ (c)\;y^2=-2x^4+25x^2 \\ (d)\;y^2=-2x^2+3x^3 $

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Let slope of tangent $\large\frac{dy}{dx}$$=m$
So we can get $AB,C,D$ writing equation of tangent and normal.
$T : y-Y=m(x-X)$
$N : y-Y=\large\frac{-1}{m}$$(x-X)$
Using section formula,
$X= \large\frac{x- \Large\frac{Y}{m}}{1+ \Large\frac{1}{\lambda}}$
$\large\frac{1}{\lambda}$$+1=1- \large\frac{Y}{mX}$
$\large\frac{1}{\lambda} =\large\frac{-Y}{mX}$
$\large\frac{AO}{OB}$$=\lambda=\large\frac{-mX}{Y}$
Similarly we can get,
$\large\frac{OC}{CD}$$=\large\frac{-mY}{X+mY}$
$\large\frac{AO}{OB}=\large\frac{OC}{CD}$
$\large\frac{mx}{-Y}=\large\frac{-mY}{X+mY}$
$y^2-x^2=mXY$
$\large\frac{dy}{dx}$$.x.y=y^2-x^2$
Let $y=vx$
$\bigg( x \large\frac{dv}{dx}$$+v\bigg)=\large\frac{-1}{V}$
$-vdv=xdx$
$\large\frac{v^2}{2}=\frac{-x^2}{2}$$+c$
$\large\frac{y^2}{2x^2}=\frac{-x^2}{2}$$-c$
(It satisfy 2,3)
So required curve is
$y^2=-x^4+\large\frac{25}{4}$$x^2$
Hence b is the correct answer.
answered Feb 10, 2014 by meena.p
 

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