Given a general parabola in parametric form $$\big(at^2+bt, ct^2+dt\big)$$ what are the equations of the axis of symmetry, as well as the tangent at the vertex?

Using standard but rather tedious algebraic expansion, the following can be worked out:

Cartesian form:
(by eliminating $t$) $$(ax-cy)^2=(bc-ad)(dy-bx)$$

Axis of symmetry:
(using the solution here) $$ax-cy+\frac {(ab+cd)(bc-ad)}{2(a^2+c^2)}=0$$

Tangent at vertex:
(by first deriving vertex coordinates as intersection between axis of symmetry and parabola, and then forming equation of line through it with slope perpendicular to axis of symmetry) $$cx+ay-\frac {(ab+cd)^2}{4(a^2+c^2)}=0$$

Desmos implementation here.

However it would be interesting to see if there are other methods of arriving at these results quickly and elegantly, perhaps by using vector or matrix methods. Any other insights, geometric or otherwise, which can be derived from the final form would also be very helpful. The terms look very much like dot products of vectors as well as matrix determinants.

  • $\begingroup$ Very quick question... How would you find the axis of symmetry for a tilted parabola. It works well (thanks for the Desmos), but just wondering, what steps did you take? $\endgroup$ – Christopher Marley May 5 '18 at 18:28
  • $\begingroup$ @ChristopherMarley - See edits for further information. Thanks for your comments. $\endgroup$ – hypergeometric May 5 '18 at 18:39
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    $\begingroup$ @Blue - See edits for further information. Thanks for your comments. $\endgroup$ – hypergeometric May 5 '18 at 18:39

Define points $P$ and $Q$ at $t=p$ and $t=q$. Invoking the calculus, we know that tangent vectors at $P$ and $Q$ are $$P^\prime=(2ap+b,2cp+d) \qquad Q^\prime = (2aq+b,2cq+d)$$ Now, suppose $P^\prime$ and $Q^\prime$ are orthogonal. Then $$0 = P^\prime \cdot Q^\prime = b^2 + d^2 + 2 (p+q)( a b + c d ) + 4 (a^2+c^2) p q$$ so that $$q = -\frac{b^2 + d^2 + 2 p(a b +cd)}{2 ( a b + cd + 2p( a^2 + c^2))}$$ unless the denominator is zero. That is, if $$p = -\frac{ab+cd}{2(a^2+c^2)} \tag{$\star$}$$ then there's no vector $Q^\prime$ orthogonal to $P^\prime$. (Ignoring possible degeneracies in the parameters $a$, $b$, $c$, $d$,) This can only happen if $P$ is the vertex.

So, substituting $p$ from $(\star)$ into $P$ and $P^\prime$ gives the vertex and tangent-vector-at-vertex for the parabola. The equations for the axis of symmetry and tangent line are straightforward to derive from this information. The cleanup remains a bit tedious, but the work is complete. $\square$

  • $\begingroup$ Very nice solution. Thanks. (+1) $\endgroup$ – hypergeometric May 6 '18 at 3:45

Given $$ \left\{ \matrix{ x = at^{\,2} + bt \hfill \cr x' = 2at + b \hfill \cr y = ct^{\,2} + dt \hfill \cr y' = 2ct + d \hfill \cr} \right. $$

then the inclination of the axis will be given by the limit, in our case for $t \to \infty$, of the ratio $y/x$ $$ y - y_v = m\left( {x - x_v } \right)\;:\quad m = \mathop {\lim }\limits_{t \to \infty } {y \over x} = \mathop {\lim }\limits_{t \to \infty } {{y'} \over {x'}} = {c \over a} $$

At the vertex, the inclination of the tangent shall be normal to the axis, i.e. $$ {{y'(t_v )} \over {x'(t_v )}} = - {1 \over m} = - {a \over c} $$

from which we find the value $t=t_v$ for the parameter at the vertex $$ \eqalign{ & 2c^2 t_v + cd = - 2a^2 t_v - ab \cr & t_v = {{ - ab - cd} \over {2\left( {a^2 + c^2 } \right)}} \cr} $$

Then, to determine the aperture, and depending on the goals of your study, you can either
- calculate the radius of curvature, which for a parametric curve is $$ R = \left| {{{\left( {x'^2 + y'^2 } \right)^{3/2} } \over {x'y'' - x''y'}}} \right| $$ and that in our case for $t=t_v$ becomes $$ R_{\,v} = {{\left( { - ad + bc} \right)^2 } \over {2\left( {a^2 + c^2 } \right)^{3/2} }} $$ - or just take another point at a $t \ne t_v$

then you have all the data to construct the normal equation of the parabola.

  • $\begingroup$ Very nice solution. Thanks. (+1) $\endgroup$ – hypergeometric May 6 '18 at 3:45
  • $\begingroup$ How does the radius of curvature help in deriving the axis of symmetry and tangent at vertex? $\endgroup$ – hypergeometric May 6 '18 at 16:49
  • $\begingroup$ @hypergeometric: axis, tangent at v. and vertex are already determined in the first part. R is to help in determining the "aperture": in the parabola $y=kx^2$ the radius is $R=1/y''=1/(2k)$. So taking $\bf n$ and $\bf t$ as the unit vectors along the axis and the tangent you can write $\mathop {VP}\limits^ \to \cdot {\bf n} = 1/\left( {2R} \right)\left( {\mathop {VP}\limits^ \to \cdot {\bf t}} \right)^{\,2} $ $\endgroup$ – G Cab May 6 '18 at 18:15

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