Parabola tangent to two lines and through two points on those lines

Question

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.

Answer 1

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

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.