Suppose the equation of axis of symmetry of a parabola is $a_1x+b_1y+c_1=0$ and the equation of the line perpendicular to the axis of symmetry and passing through the vertex $a_2x+b_2y+c_2=0$. I noticed while doing a problem that the equation of the parabola turned out to be $$(a_1x+b_1y+c_1)^2=k(a_2x+b_2y+c_2)$$ But I could'nt understand what 'k' was. Please help.

  • $\begingroup$ forgot to mention it in the the question ; 'k' is a constant $\endgroup$ – Gautham Mohan Aug 26 '16 at 13:41
  • $\begingroup$ There are infinite parabolas and $k$ is a parameter. Consider the simple case when the axis of symmetry is $x=0$ and the second line is $y=0$, then the parabolas are $x^2=ky$ for any $k\not=0$. $\endgroup$ – Robert Z Aug 26 '16 at 13:47
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    $\begingroup$ The focus is not specified yet and it's in turn governed by $k$. $\endgroup$ – Ng Chung Tak Aug 26 '16 at 13:50
  • $\begingroup$ if the focus to be (m,n) then is there any way we can find out the value of k $\endgroup$ – Gautham Mohan Aug 26 '16 at 13:53
  • $\begingroup$ en.wikipedia.org/wiki/Parabola#General_parabola $\endgroup$ – Intelligenti pauca Aug 26 '16 at 15:28

Some observations first. Since your two lines are perpendicular to one another, the two sides of your quation are essentially like $x^2=ky$, except that you applied some coordinate transformation to a different affine coordinate system whose basis is orthogonal but not neccessarily orthonormal. The orthogonality of the lines can be verified by $a_1a_2+b_1b_2=0$. Since the equation $(ka_2)x + (kb_2y) +(kc_2)$ is essentially of the same form as $a_2x+b_2y+c_2$, you might as well speak about the parabola being defined by the two lines alone.

Finding focus from equation

Judging from your comment, it looks as though your main goal is trying to determine the location of the focus if you have all the parts of the equation.

So let's concentrate on $x^2=ky$ first. Where is the focus here? Well, $y=x^2$ has the focus at $(0,\frac14)$. If you scale the $y$ side of the equation by $k$ you get the focus at $ky=\frac14$ or $y=\frac1{4k}$. This covers the case of $a_1=1,b_1=0,c_1=0,a_2=0,b_2=1,c_2=0$. If you want $a_1$ and $b_2$ to be arbitrary lengths you get the equation $(a_1x)^2=kb_2y$ and the focus at $(0,\frac{a_1^2}{4kb_2})$.

Next you consider $b_1\neq0$ and $a_2\neq0$ while keeping $c_1=c_2=0$ for the moment. That's essentially a rotation. The distance from the origin (which is still the vertex) to the focus can be derived similar to the equation above, except that instead of $a_1$ and $b_2$ you now should use the length of the corresponding vectors. So if the distance between vertex and focus were $d$ you'd have


The direction from the vertex to the focus is the direction indicated by $(a_2,b_2)$, but you have to normalize that to unit length. So the vector from vertex to focus is


In order to also consider $c_1\neq0$ and / or $c_2\neq0$ all you need are coordinates for the focus. You can compute them in homogeneous coordinates using the cross product.

$$\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix} \times\begin{pmatrix}a_2\\b_2\\c_2\end{pmatrix} =\begin{pmatrix} b_1c_2-c_1b_2\\ c_1a_2-a_1c_2\\ a_1b_2-b_1a_2 \end{pmatrix}$$

which dehomogenizes to

$$V=\frac{1}{a_1b_2-b_1a_2} \begin{pmatrix}b_1c_2-c_1b_2\\c_1a_2-a_1c_2\end{pmatrix}$$

so you have

$$F=V+\overrightarrow{VF}= \frac{1}{a_1b_2-b_1a_2} \begin{pmatrix}b_1c_2-c_1b_2\\c_1a_2-a_1c_2\end{pmatrix}+ \frac{a_1^2+b_1^2}{4k\left(a_2^2+b_2^2\right)}\begin{pmatrix}a_2\\b_2\end{pmatrix}$$

as the coordinates of the focus, depending on $k$ and the other coefficients of your lines.

Finding equation from axis, vertex and focus

The title of the question duggests a different problem, though. Namely that of finding the equation of the equation given the vertex and the focus. The axis of symmetry is alreasdy implied by these two points. So let's suppose you have your vertex at $V=(V_x,V_y)$ and your focus at $F=(F_x,F_y)$. Then the vector $4\overrightarrow{VF}$ plays the same role for this parabola as the vector $(0,1)$ plays for the standard parabola $y=x^2$. If that vector has length one, you can extract the analogon of the $y$ coordinate from a point $P=(x,y)$ by computing the dot product $4(F-V)\cdot(P-V)$. Otherwise you have to divide by the squared length of that vector: $\frac{4}{\lVert F-V\rVert^2}(F-V)\cdot(P-V)$. That's essentially the right hand side of your equation.

For the left hand side you need the same vector $F-V$ but rotated by $90°$ in either direction. So let's make that $(V_y-F_y,F_x-V_x)$. Thus the complete equation would be

\begin{multline*} \left(\frac{1}{(F_x-V_x)^2+(F_y-V_y)^2}\bigl((V_y-F_y)(x-V_x)+(F_x-V_x)(y-V_y)\bigr)\right)^2=\\=4\left(\frac{1}{(F_x-V_x)^2+(F_y-V_y)^2}\bigl((F_x-V_x)(x-V_x)+(F_y-V_y)(y-V_y)\bigr)\right) \end{multline*}

You can multiply by those denominators and cancel common factors.

\begin{multline*} \bigl((V_y-F_y)(x-V_x)+(F_x-V_x)(y-V_y)\bigr)^2=\\=4\bigl((F_x-V_x)(x-V_x)+(F_y-V_y)(y-V_y)\bigr)\bigl((F_x-V_x)^2+(F_y-V_y)^2\bigr) \end{multline*}

Now you can easily expand these terms to read $a_1,b_1,c_1,a_2,b_2,c_2$ from these. As I write above, you can essentially choose the $k$ any way you like (except for $k=0$).

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