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Starting from the parabola

$$ y = a x^2 $$

I want to draw it after tilting by an angle $\theta$ clockwise such that it is tangent to the positive $x$ axis and the positive $y$ axis.

To be able to draw it I need to find the location of the vertex.

Question: How can I find the vertex coordinates ?

My effort:

I've parameterized the parabola as the vector p(t) given by

$ p(t) = V + \ t \ u_1 + a t^2 \ u_2 $

where $ u_1 = [ \cos \theta , - \sin \theta ]^T $ and $ u_2 = [ \sin \theta , \cos \theta ]^T $

The tangent vector to $p(t)$ is its derivative

$p'(t) = u_1 + 2 a t \ u_2 $

At the tangency point with the $x$ axis, the dot product of $p'(t)$ with $[0, 1]^T$ is zero.

Hence,

$ 0 = (0, 1) \cdot ( u_1 + 2 a t_x \ u_2 ) = u_{1y} + 2 a t_x \ u_{2y} $

From which

$ t_x = - \dfrac{1}{2 a} \dfrac{ u_{1y} }{u_{2y} } = \dfrac{1}{2a} \tan \theta $

Similarly, at the tangency point with the $y$ axis, the dot product of $p'(t)$ with [1, 0]^T $ is zero.

Hence,

$ 0 = (1, 0) \cdot ( u_1 + 2 a t_y \ u_2 ) = u_{1x} + 2 a t_y \ u_{2x} $

From which

$ t_y = - \dfrac{1}{2a} \dfrac{ u_{1x} }{u_{2x} } = - \dfrac{1}{2 a} \cot \theta $

Next, the $y$ coordinate of the tangency point with $x$ axis is zero. Hence,

$ 0 = V_y + t_x u_{1y} + a t_x^2 u_{2y} = V_y - \dfrac{1}{2a} \tan \theta \sin \theta + a \left( \dfrac{1}{2a} \tan \theta \right)^2 \cos \theta $

And this simplifies to

$ 0 = V_y - \dfrac{1}{4a} \dfrac{ \sin^2 \theta }{ \cos \theta } $

From which

$ V_y = \dfrac{1}{4a} \dfrac{ \sin^2 \theta }{ \cos \theta } $

Similarly, the $x$ coordinate of the tangency point with $y$ axis is zero. Hence,

$ 0 = V_x + t_y \ u_{1x} + a t_y^2 \ u_{2x} = V_x - \dfrac{1}{2a} \cot \theta \cos \theta + \dfrac{1}{4 a} \cot^2 \theta \sin \theta $

And this simplifes to

$ V_x = \dfrac{1}{4a} \dfrac{\cos^2 \theta} {\sin \theta } $

Now that the vertex coordinates are known, I can draw my parabola.

Question Recap: The above method works, but is there a way to find the vertex coordinates using the algebraic equation of the parabola instead of the parametric equation ?

The algebraic equation is

$ (r - V)^T R D R^T (r - V) + b_0^T R^T (r - V) = 0 $

where $V$ is the unknown vertex, and

$ r = [x, y]^T $

$ D = \begin{bmatrix} a && 0 \\ 0 && 0 \end{bmatrix}$

$ R = \begin{bmatrix} \cos \theta && \sin \theta \\ - \sin \theta && \cos \theta \end{bmatrix} $

(note that rotation here is clockwise)

$ b_0 = [0, -1]^T $

I wasn't able to derive the vertex coordinates starting from the parabola's algebraic equation.

Hosam Hajeer
  • 21,978

1 Answers1

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Starting with the parabola $$\pm\sqrt{\frac{x}{p}}\pm\sqrt{\frac{y}{q}}=1,$$ which touches the axes at $(p,0),(0,q),$ we can get

$$\frac{(Bx+Ay-\frac{AB}{A^2+B^2})}{\sqrt{A^2+B^2}}=a\frac{(Ax-By+\frac{B^2-A^2}{A^2+B^2})^2}{A^2+B^2},\\ A=\frac1{p}> 0, B=\frac1{q}> 0, a=\frac{(A^2+B^2)^\frac32}{4AB},$$ that is with $A,B$ on the first quadrant leaf of the curve $a^216A^2B^2=(A^2+B^2)^3$ (or in polar coordinates $r=2a\sin{2\theta}.$) Then the vertex is $(\frac{A^3}{(A^2+B^2)^2},\frac{B^3}{(A^2+B^2)^2})=(\frac{\cos^3{\theta}}{2a\sin{2\theta}},\frac{\sin^3{\theta}}{2a\sin{2\theta}})$ and the equation becomes $$x\sin{\theta}+y\cos{\theta}-\frac1{4a}=a(x\cos{\theta}-y\sin{\theta}-\frac{\cot{2\theta}}{2a})^2.$$

Jan-Magnus Økland
  • 5,149