$(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.

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