Study of parabolas of the form $(ax + by) ^ 2$ + $2gx + 2fy + c = 0$

Question

I am studying parabolas of the form $(ax + by) ^ 2$ + $2gx + 2fy + c = 0$. In the text which I am referring to it is mentioned that we can do the following manipulations:

$$(ax + by) ^ 2 = -2gx - 2fy - c$$ We now add an arbitrary constant $\alpha$ in the square root of the second degree terms. Then the equation transforms to: $$(ax + by + \alpha) ^ 2 = xf_1(\alpha) + yf_2(\alpha) + f_3(\alpha)$$

We choose $\alpha$ such that the lines $ax + by + \alpha = 0$ and $xf_1(\alpha) + yf_2(\alpha) + f_3(\alpha) = 0$ are perpendicular.

My questions is why should the lines $ax + by + \alpha = 0$ and $xf_1(\alpha) + yf_2(\alpha) + f_3(\alpha) = 0$ be perpendicular ?

Answer 1

A general parabola equation is of the type $y=4ax^2$in your question we want to compare both the equations such that $ax+by +\alpha $ like y axis and $xf_1(\alpha) is +yf_2(alpha) +f_3(alpha) $ is like x axis thus we will be able to use equations derived for $y^2=4ax$ here also and so $ax+by +\alpha $ and $xf_1(\alpha) is +yf_2(alpha) +f_3(alpha) $ has to be perpendicular

Answer 2

I suspect that the reason will be revealed later in the text. With a suitable choice of $\alpha$ (a bad choice of symbol, by the way, since you’re already using $a$, which is too similar-looking), the quantity being squared on the left-hand side gives you an equation of the parabola’s axis. The perpendicular line on the right-hand side of the parabola’s transformed equation is its tangent at the vertex, which is of course perpendicular to the axis. The vertex of the parabola is the intersection of these two lines.

Taking, for example, the parabola $(3x+4y)^2+2x-4y+5=0$, we want the left-hand side to be $(3x+4y+\lambda)^2$. Subtracting the original left-hand side from this gives $(6\lambda-2)x+(8\lambda+4)y+(\lambda^2-5)$. We want this to be some multiple of $4x-3y+\mu$. This latter constraint can be written as $$\begin{vmatrix}6\lambda-2 & 8\lambda+4 \\ 4 & -3 \end{vmatrix} = -50\lambda-10 = 0,$$ from which $\lambda = -\frac15$, and so the original equation can be rewritten as $$\left(3x+4y-\frac15\right)^2 = -{16\over5}x+{12\over5}y-{124\over25}.$$ If you plot the parabola and the two lines derived from the last equation, you’ll see that they are indeed the parabola’s axis and tangent at the vertex.

A different way to see what’s going on is to start from the standard equation $y^2=4px$. Hiding in this equation are the equations of the axis and vertex tangent. This parabola’s axis is the $x$-axis, which has equation $y=0$, and the tangent is the $y$-axis, with equation $x=0$. A rotation and translation is effected by making the substitutions $x\to x'\cos\theta+y'\sin\theta+h$ and $y\to -x'\sin\theta+y'\cos\theta+k$, producing $$(-x'\sin\theta+y'\cos\theta+k)^2=4p(x'\cos\theta+y'\sin\theta+h),$$ but $-x'\sin\theta+y'\cos\theta+k=0$ is the image of the $x$-axis, and similarly, $x'\cos\theta+y'\sin\theta+h=0$ is the image of the $y$-axis. You can also include a nonzero scale factor $C$, i.e., $$(-Cx'\sin\theta+Cy'\cos\theta+Ck)^2=4C^2p(x'\cos\theta+y'\sin\theta+h).$$ Applying this idea to the above example, we normalize the left-hand side by pulling out a factor of $3^2+4^2=25$ and pull out a corresponding factor on the right to make the coefficients of the two parenthesized expressions match: $$\left(\frac35x+\frac45y-\frac1{25}\right)^2 = \frac4{25} \left(-{4\over5}x+{3\over5}y-{31\over25}\right).$$ From this equation we can read the parabola’s parameter $p=1/25$.