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Given a general parabola $$(Ax+Cy)^2+Dx+Ey+F=0$$ what is the axis of symmetry in the form $ax+by+c=0$?

It is possible of course to first work out the angle of rotation such that $xy$ and $y^2$ terms disappear, in order to get an upright parabola $y=px^2+qx+r$ and proceed from there. This may involve some messy trigonometric manipulations.

Could there be another approach perhaps, considering only quadratic and linear equations?

Addendum

From the solution (swapped) by Meet Taraviya and some graphical testing, the equation for the axis of symmetry is

Axis of Symmetry: $$\color{red}{Ax+Cy+\frac {AD+CE}{2(A^2+C^2)}=0}$$

which is quite neat. Note that the result is independent of $F$. Awaiting further details on the derivation.

Addendum 2

Here is an interesting question on MSE on a similar topic.

Addendum 3 (added 23 May 2018)

Tangent at Vertex: $$Cx-Ay+\frac {(A^2+C^2)(F-k^2)}{CD-AE}=0$$ where $k=\frac {AD+CE}{2(A^2+C^2)}$.

Note that the parabola can also be written as

$$\underbrace{Cx-Ay+d}_{\text{Tangent at Vertex if $=0$}} =m\;\big(\underbrace{Ax+Cy+k}_{\text{Axis of Symmetry if $=0$}}\big)^2$$ where $$m=\frac {A^2+C^2}{AE-CD}$$ and $d=\frac {(A^2+C^2)(F-k^2)}{CD-AE}$

See Desmos implementation here.

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Write the equation as:- $$(Ax+Cy+t)^2+(D-2At)x+(E-2Ct)y+F-t^2=0$$ Choice of t is made such that $A\cdot(D-2At)+C\cdot(E-2Ct)=0$

$Ax+Cy+t=0$ is the symmetry axis of parabola.

Also, the line $(D-2At)x+(E-2Ct)y+F-t^2=0$ is the tangent at vertex of the parabola.

Explanation :-

Interpret $y^2=4ax$ as:- $$(Distance \space from \space x=0)^2=4a\cdot (Distance \space from \space y=0)$$

Note that $y=0$ is the symmetry axis of the parabola and $x=0$ is the tangent at vertex. Also, they are perpendicular to each other (This explains why $A\cdot(D-2At)+C\cdot(E-2Ct)=0$ must be true for these lines to be them. This statement is equivalent to $m_1m_2=-1$)This property holds true for a general parabola.

Thus a parabola can be represented as:-

$$(Distance \space from \space L_1)^2=4a\cdot (Distance \space from \space L_2)$$ where $L_1$ and $L_2$ are the symmetry axis and the tangent at vertex of perpendicular.

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  • 1
    $\begingroup$ Thanks - very nice solution! (+1) $\endgroup$ – hypergeometric Dec 6 '16 at 18:00

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