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is it possible to calculate parabola that is tangent to two lines exactly on black points? (please see enclosed picture)

And linked question is:

If we assume red line is given by: $$ f_{1}(x)=S_{1}(x-A)+B $$

And green one is: $$ f_{2}(x)=S_{2}(x-C)+D $$

Then what requirements I need to meet with $A$, $B$, $C$, $D$, $S_1$, $S_2$ to make my task possible?

For any help great thanks in advance.

enter image description here

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  • $\begingroup$ Effectively a duplicate of math.stackexchange.com/q/2622143/265466. See also math.stackexchange.com/q/2999165/265466. $\endgroup$ – amd Mar 15 at 7:54
  • $\begingroup$ Given two distinct lines with one point on each line (we can't have the intersection point of the two lines), we can in the general parabola equation $ax^2+2bxy+cy^2+2dx+2ey+f=0, b^2-ac=0$ find the $a,b,c,d,e,f$ up to a scalar factor from your data. We find four equations in addition to $b^2-ac=0$, two from the conditions for the points to be on the conic, two more from the points dual to the lines lying on the dual of the parabola. The solutions will be the double line through the two points (since this technically is tangent to anything), the all zero solution, and the equation sought. $\endgroup$ – Jan-Magnus Økland Mar 15 at 9:58
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In sage, using the trick of differentiating from the answer by @stevengregory:

var("a,b,d,e,f,A,B,C,D,S1,S2")
eq1=(a*A+b*B)^2+d*A+e*B+f==0
eq2=(a*C+b*D)^2+d*C+e*D+f==0
eq3=2*(a*A+b*B)*(a+b*S1)+d+e*S1==0
eq4=2*(a*C+b*D)*(a+b*S2)+d+e*S2==0
solve([eq1,eq2,eq3,eq4],a,b,d,e,f)

$(\frac{BC-AD-(B-D)x+(A-C)y}{B-D})^2=0$

As you can see you get the double line through the two points. This is because technically a double line is tangent where it intersects.

$(ax+(-(A(S_1 + S_2) - C(S_1 + S_2) - 2B + 2D)a/(2AS_1S_2 - 2CS_1S_2 - B(S_1 + S_2) + D(S_1 + S_2)))y)^2+(-2(2A^3S_1^2S_2^2 + 2C^3S_1^2S_2^2 - 2(2S_1^2S_2 + S_1S_2^2)C^2D + 2(S_1^2 + 2S_1S_2)CD^2 - 2D^3S_1 - 2B^3S_2 - 2(CS_1^2S_2^2 - DS_1S_2^2)A^2 + (2(2S_1S_2 + S_2^2)A + (S_1^2 - 2S_1S_2 - S_2^2)C - 2D(S_1 - 2S_2))B^2 - (2C^2S_1^2S_2^2 - 4CDS_1^2S_2 + (S_1^2 + 2S_1S_2 - S_2^2)D^2)A + (2C^2S_1^2S_2 - 2(S_1^2S_2 + 2S_1S_2^2)A^2 - (3S_1^2 + 2S_1S_2 - S_2^2)CD + 2D^2(2S_1 - S_2) + (4CS_1S_2^2 + (S_1^2 - 2S_1S_2 - 3S_2^2)D)A)B)a^2/(4A^2S_1^2S_2^2 + 4C^2S_1^2S_2^2 + (S_1^2 + 2S_1S_2 + S_2^2)B^2 - 4(S_1^2S_2 + S_1S_2^2)CD + (S_1^2 + 2S_1S_2 + S_2^2)D^2 - 4(2CS_1^2S_2^2 - (S_1^2S_2 + S_1S_2^2)D)A - 2(2(S_1^2S_2 + S_1S_2^2)A - 2(S_1^2S_2 + S_1S_2^2)C + (S_1^2 + 2S_1S_2 + S_2^2)D)B))x+(2(2A^3S_1^2S_2 + 2C^3S_1S_2^2 - 2(2S_1S_2 + S_2^2)C^2D + 2CD^2(S_1 + 2S_2) - (2(2S_1^2S_2 - S_1S_2^2)C - (S_1^2 + 2S_1S_2 - S_2^2)D)A^2 + 2(A(2S_1 + S_2) - CS_1 + D)B^2 - 2B^3 - 2D^3 + (2(S_1^2S_2 - 2S_1S_2^2)C^2 - (S_1^2 - 2S_1S_2 - 3S_2^2)CD - 2D^2S_2)A - (2(S_1^2 + 2S_1S_2)A^2 + (S_1^2 - 2S_1S_2 - S_2^2)C^2 + 4CDS_2 - ((3S_1^2 + 2S_1S_2 - S_2^2)C - 4DS_1)A - 2D^2)B)a^2/(4A^2S_1^2S_2^2 + 4C^2S_1^2S_2^2 + (S_1^2 + 2S_1S_2 + S_2^2)B^2 - 4(S_1^2S_2 + S_1S_2^2)CD + (S_1^2 + 2S_1S_2 + S_2^2)D^2 - 4(2CS_1^2S_2^2 - (S_1^2S_2 + S_1S_2^2)D)A - 2(2(S_1^2S_2 + S_1S_2^2)A - 2(S_1^2S_2 + S_1S_2^2)C + (S_1^2 + 2S_1S_2 + S_2^2)D)B))y+((4(CS_1^2S_2^2 - DS_1^2S_2)A^3 - 4(CS_2 - D)B^3 - (8C^2S_1^2S_2^2 - 4(3S_1^2S_2 + S_1S_2^2)CD + (3S_1^2 + 6S_1S_2 - S_2^2)D^2)A^2 + ((S_1^2 - 6S_1S_2 - 3S_2^2)C^2 + 4CD(S_1 + 3S_2) + 4((2S_1S_2 + S_2^2)C - D(2S_1 + S_2))A - 8D^2)B^2 + 4(C^3S_1^2S_2^2 - (2S_1^2S_2 + S_1S_2^2)C^2D + (S_1^2 + 2S_1S_2)CD^2 - D^3S_1)A - 2(2C^3S_1S_2^2 - 2(2S_1S_2 + S_2^2)C^2D + 2CD^2(S_1 + 2S_2) + 2((S_1^2S_2 + 2S_1S_2^2)C - (S_1^2 + 2S_1S_2)D)A^2 - 2D^3 - (2(S_1^2S_2 + 3S_1S_2^2)C^2 - (3S_1^2 + 10S_1S_2 + 3S_2^2)CD + 2D^2(3S_1 + S_2))A)B)a^2/(4A^2S_1^2S_2^2 + 4C^2S_1^2S_2^2 + (S_1^2 + 2S_1S_2 + S_2^2)B^2 - 4(S_1^2S_2 + S_1S_2^2)CD + (S_1^2 + 2S_1S_2 + S_2^2)D^2 - 4(2CS_1^2S_2^2 - (S_1^2S_2 + S_1S_2^2)D)A - 2(2(S_1^2S_2 + S_1S_2^2)A - 2(S_1^2S_2 + S_1S_2^2)C + (S_1^2 + 2S_1S_2 + S_2^2)D)B))=0$

The other solution is the unique parabola from your data.

For example for

A=-2
B=3
C=3
D=2
S1=-2
S2=1/2

we get

ABCDS1S2

The problem is a degenerate case of the four conic sections through two points tangent to three lines in general position in ${\Bbb P}^2_{\Bbb C}$. The two points and two lines you supply are not in general position, since the points lie on the lines. But, since you seek a parabola, the line at infinity; $z=0$ is a third tangent line.

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