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In this answer, a point P is considered on the conic, the conic is shifted by point P , and it turns out that the linear part of this shifted equation denotes the 'shifted' equation of tangent at P. So, unshifting linear part gives equation of tangent of the conic. This was supposed to be intuitive, but I just don't get why it should happen that the linear part of shifted conic represents of the shifted conci is shifted tangent line without working it out through other methods.

Any hints is appreciated

tryst with freedom
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    Effectively, this says that the linear part of the equation of a conic through the origin gives the tangent line at the origin. Eg, for $Ax^2+Bxy+Cy^2+Dx+Ey=0$, the tangent line at the origin is $Dx+Ey=0$. One can do the calculus thing by taking appropriate derivatives and substituting in $x=y=0$ to find $D+Ey'=0$, and the result follows. Less-formally, one can make the common argument that "near $0$", higher-power terms (here, the quadratics) "vanish faster", becoming negligible, leaving the linear terms to describe the infinitesimal behavior of the curve at the origin. – Blue Jan 30 '21 at 20:38
  • Hey @Blue, that was a nice idea for understanding it. I think I get it now :^) – tryst with freedom Jan 31 '21 at 05:43

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