find the position and Nature of the double points of the curve x2- 2ay3- 3a2y2-2a2x2+a2 =0
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Let [math]f(x,y)[/math] represent the left-hand side. Then we have [math]\dfrac{\partial f}{\partial x}=4x^3-4a^2x[/math] [math]\dfrac{\partial f}{\partial y}=-6ay^2-6a^2y[/math] The former vanishes for [math]x=0,a,-a[/math], the latter for [math]y=0,-a[/math] Assuming [math]a e0[/math], we need to compute [math]f(0,0)=a^4 e0[/math] [math]f(0,-a)=2a^4-3a^4+a^4=0[/math] [math]f(a,0)=a^4-2a^4+a^4=0[/math] [math]f(a,-a)=a^4+2a^4-3a^4-2a^4+a^4 e0[/math] [math]f(-a,0)=a^4-2a^4+a^4=0[/math] [math]f(-a,-a)=a^4+2a^4-3a^4-2a^4+a^4 e0[/math] Hence the singular points are [math](0,-a)[/math], [math](a,0)[/math] and [math](-a,0)[/math]. Let’s compute the second order derivatives [math]\dfrac{\partial^2f}{\partial x^2}=12x^2-4a^2[/math] [math]\dfrac{\partial^2f}{\partial x\,\partial y}=0[/math] [math]\dfrac{\partial^2f}{\partial y^2}=-12ay-6a^2[/math] So the Hessian matrix is [math]H_f(x,y)=\begin{bmatrix} 12x^2-4a^2 & 0 \ 0 & -12ay-6a^2 \end{bmatrix}[/math] We have then [math]H_…
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