Hi everyone can you please help me with this question? Is there no shorter way to do this then my approach? Is this a correct way to do it?

Determine the conic section in $\mathbb{R}P^2$ that is tangent to $x_2=0$ in $[(1,0,0)]$, tangent to $x_0=0$ in $[(0,-1,2)]$ and contains $[(1,0,-2)]$.

This is my approach to this problem: A conic section in $\mathbb{R}P^2$ is of the form $ax_0^2 + bx_1^2 + cx_2^2 + dx_0x_1 + ex_0x_2 + fx_1x_2$. I will now try to determine the coefficients $a ... f$. We know three points contained by the conic section: $[(1,0,0)], [(0,-1,2)],[1,0,-2)]$. This learns us:

$a+d+e=0$, $b + 4c -2f = 0$, $a + 4c -2e=0$.

The tangtent lines are of course of the form $x_0 \frac{\partial f(p)}{\partial x_0} + x_1\frac{\partial f(p)}{\partial x_1}+ x_2\frac{\partial f(p)}{\partial x_2}=0$. So now lets fill in what we know: $x_0= x_0 \frac{\partial f((1,0,0))}{\partial x_0} + x_1\frac{\partial f((1,0,0))}{\partial x_1}+ x_2\frac{\partial f((1,0,0))}{\partial x_2}=0$. So $\frac{\partial f((1,0,0))}{\partial x_1}=0$, $\frac{\partial f((1,0,0))}{\partial x_2}=0$. Do the same for the other tangent line. And then solve the linear system that we obtain.


It is correct (apart from the numbers: I guess you would like the tangent in $[1,0,0]$ to be $x_2=0$ and not $x_0=0$,which instead should be the tangent in $[0,-1,2]$).

If you know what a pencil of conics is, you could go like this: take $r_1$ to be the line through $[1,0,0]$ and $[1,0,-2]$ ($x_1=0$) and, together with the desired tangent in $[0,-1,2]$, write a degenerate conic $$C_1=\{x_0x_1=0\}\;.$$ Now do the same with the two lines through $[0,-1,2]$ and $[1,0,0]$ and $[1,0,-2]$ respectively ($2x_1+x_2=0$ and $2x_0+2x_1+x_2=0$) obtaining $$C_2=\{(2x_1+x_2)(2x_0+2x_1+x_2)=4x_1^2+x_2^2+4x_0x_1+2x_0x_2+4x_1x_2=0\}$$ and now consider the generic conic of the pencil $$C_k=\{4x_1^2+x_2^2+(4+k)x_0x_1+2x_0x_2+4x_1x_2=0\}\;.$$ Imposing that the tangent of $C$ in $[1,0,0]$ should be $x_2=0$, you are done: take the derivatives in $[1,0,0]$, which are $0, 4+k, 2$ and equate them to $[0,0,1]$ to get $k=-4$. Hence your conic should be $$C=\{4x_1^2+x_2^2+2x_0x_2+4x_1x_2=0\}\;.$$

  • $\begingroup$ $x_2=0$ should be tangent in $[(1,0,0)]$ and $x_0$ should be tangent to $[(0,-1,2)]$. I have edited it in my question as well. $\endgroup$ – Leo Aug 31 '13 at 10:47
  • $\begingroup$ yes, i wrote the answer supposing that it was as you said in this previous comment $\endgroup$ – wisefool Aug 31 '13 at 10:50

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