Determine conics by four points and a tangent line.

I have small troubles determining conics that go through four points and have a given tangent line. More specifically $P_1 = (0:1:0), P_2 = (0:0:1), P_3 = (1:0:1), P_4 = (1:-1:0) \in \mathbb{RP}^2$ and $t: \{x-y+z=0\}$.

I can determine the conics as far as $ax^2+axy+2eyz-axz=0,\ a,e \in \mathbb{R}$ but don't really succeed with incorporating the information of the tangent line.

I read substituting in $z = -x +y$ should work, but i have no clue how to proceed with the obtained result $ax^2+ey^2-exy=0$.

Help is appreciated a lot (also other ways of dealing with the tangent!).

• Assuming your algebra is correct, at points $(x:y:z)$ where $ax^2+ey^2-exy=0$ and $x-y+z=0$, you want the gradient of $ax^2+axy+2eyz-axz$ to be parallel to the gradient of $x-y+z$, i.e., $(1,-1,1)$. – Ted Shifrin Jun 11 '18 at 17:14
• @TedShifrin could you further elaborate how to achieve both gradients to be parallel? The gradient of $ax^2+axy+2eyz-axz$ still depends on $x,y,z$ - so i need the points of intersection of the tangent with the conic. But for that i need the values for $a,e$. This is what blocks me at the moment. – user526159 Jun 11 '18 at 17:28

The conic will be tangent to the line if you get a double point when you try to solve for their intersection, and the conic is smooth at that point.

Given $ax^2+axy+2eyz-axz = 0$ and $x-y+z = 0$, replacing $y$ with $x+z$ gives you $2ax^2+2exz+2ez^2 = 0$.
There is a double solution if and only if the discrimnant of the quadratic is zero, so you want
$(2e)^2 = 4(2a)(2e)$.

This is equivalent to $e(e-4a) = 0$.

If you pick $e=0$ you get $x(x+y-z) = 0$, which unfortunately is the reunion of two lines. The "tangent line" goes through their intersect point, so the conic is not smooth there, and it's not a tangent line after all.

If you pick $e=4a$ you get the hyperbola $x^2 + xy - xz + 8zy = 0$

• A quick check in geogebra for $z=1$, $y=x+1$ is not tangent to your $x^2+xy-x+4y=0$ – Jan-Magnus Økland Jun 11 '18 at 19:47
• I forgot the coefficient of $yz$ was $2e$ and not $e$ ... thanks for noticing ! – mercio Jun 11 '18 at 20:24
• @mercio Thank you! – user526159 Jun 11 '18 at 20:51

$\det(\begin{pmatrix}x^2&xy&y^2&xz&yz&z^2\\0&0&1&0&0&0\\0&0&0&0&0&1\\1&0&0&1&0&1\\1&-1&1&0&0&0\\a^2&ab&b^2&ac&bc&c^2\end{pmatrix})=a(c-b-a)yz+bc(-xz+xy+x^2)=0$

Now assume $(a:b:c)$ is the point of tangency making $b=a+c$ and $y=x+z$, substitute and expand:

$2((ac+c^2)x^2-a^2xz-a^2z^2)$

The discriminant reduces to $a^2(2c+a)^2$, so $a=-2c,b=-c,c=c$, which makes the equation $-c^2(8yz-xz+xy+x^2)$, and this with $z=1$ is a hyperbola that actually is tangent to $y=x+1$ at $(-2,-1)$.

• Thank you! I like the clean look of the determinant, but didn't chose your answer because i basically only missed the fact that the discriminant has to be zero and also because this (at least for me) takes longer to compute manually. – user526159 Jun 11 '18 at 21:30
• is $a(c-b-a)xz$ supposed to be $a(c-b-a)yz$ ? – mercio Jun 12 '18 at 7:57
• @mercio: nice catch. – Jan-Magnus Økland Jun 12 '18 at 8:35