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If a parabola touches the line $y=x$ and $y=-x$ at $A(3,3)$ and $B(1,-1)$, then find the focus, axis of the parabola and its directrix.

What I thought: Since the 2 tangents are perpendicular,the origin must lie on the directrix and the line joining A and B is a focal chord. Don't know how to proceed from here..

The most general form of a parabola is $$Ax^2+Bxy + Cy^2 + Dx +Ey +F=0$$
where $$B^2 = 4AC tag{1}$$.

Letting $(x,y)=(3,3)$, we get $$9A + 9B + 9C + 3D + 3E + F = 0 tag{2}$$

Letting $(x,y)=(1, -1)$, we get $$A - B + C + D - E + F = 0 tag{3}$$

begin{align}
Ax^2+Bxy + Cy^2 + Dx +Ey +F=0
&implies 2Ax + By + Bxy' + 2Cyy' + D + Ey' = 0 \
&implies (Bx + 2Cy + E)y' + (2Ax + By + D) = 0
end{align}

Letting $(x,y,y') = (3,3,1)$, we get $(3B + 6C + E) + (6A + 3B + D) = 0$,
which implies $$6A + 6B + 6C + D + E = 0 tag{4}$$

Letting $(x,y,y') = (1,-1,-1)$, we get $-(B - 2C + E) + (2A - B + D) = 0$,
which implies $$2A - 2B + 2C + D - E = 0tag{5}$$

Solving equations $(1)$ through $(5)$ and letting $A=1$, we get

$$(A,B,C,D,E,F) = (1, -4, 4, -12, 6, 9)$$

So, the equation of the parabola becomes

$$x^2 - 4xy + 4y^2 - 12x + 6y + 9 = 0$$

Added because of something that I found out later.

If $A=0$, then $B^2=4AC$ implies $B=0$ and the equation becomes $Cy^2 + Dx +Ey +F=0$.

If $A ne 0$, then $A < 0$ implies
$$(-A)x^2 +(-B)xy + (-C)y^2 + (-D)x + (-E)y + (-F) = 0$$
and $B^2=4AC iff (-B)^2 = 4(-A)(-C)$

So we may as well assume that $A = a^2 > 0$. Then

begin{align}
Ax^2 + Bxy + Cy^2
&= dfrac{1}{4A}(4Ax^2 + 4ABxy + B^2y^2) \
&= dfrac{1}{4A}(2Ax+By)^2 \
&= (ax+by)^2 \
end{align}

where $b = dfrac{B}{2a}$.

So we can write the most general form of a parabola as $$(ax + by)^2 + Dx +Ey +F=0$$

We can now argue much as I did above and get the same answer without being annoyed by the nonlinear equation $B^2 = 4AC$.

Let’s continue with the geometric construction that you started.

Construct the parallelogram $AOBC$. The diagonal $OC$ is parallel to the parabola’s axis. We know that perpendicular tangents meet on the directrix, which is perpendicular to the parabola’s axis, so we now know its directrix $d$. Construct circles centered on $A$ and $B$ and tangent to $d$. Their intersection is the parabola’s focus. You can simply divide $overline{AB}$ proportionally to find this point: $F = {r_BA+r_ABover r_B+r_A}$. You can then use the point-point and point-line distance formulas to obtain a Cartesian equation for the parabola.

The construction in the preceding paragraph took advantage of the perpendicularity of the two tangents, but it’s quite easy to obtain a Cartesian equation of a parabola from any pair of points $P_0$ and $P_2$ on it and the tangents at those points: Let $P_1$ be the intersection of the tangents. A Bézier parameterization of the parabola is then $(1-t)^2P_0+2t(1-t)P_1+t^2P_2$. Eliminating $t$ gets you a Cartesian equation for it. In this case, the tangents intersect at the origin, so we have the parameterization $$x = 3(1-t)^2+t^2 = 4t^2-6t+3 \ y=3(1-t)^2-t^2 = 2t^2-6t+3.$$ Eliminating $t$ produces the equation $$x^2-4xy+4y^2-12x+6y+9=0.$$