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Suppose the axis of parabola is given by $\alpha x+\beta y+\lambda=0$, and tangent at vertex is $\beta x-\alpha y+\mu=0$ , now by property of parabola the distance from directrix and from focus are equal , lets assume focus is at a distance "$a$" from the vertex tangent . Now we have that $\frac{\beta x-\alpha y+\mu+a \sqrt{\alpha^{2}+\beta^{2}}}{\left(\sqrt{\alpha^{2}+\beta^{2}}\right)^{2}}=\sqrt{(x-h)^{2}+(y-k)^{2}}$ where $(h,k)$ is focus location , how do we eliminate h,k as such to get the general form of parabola from here ? As such only equation relating $(h,k)$ is the axis line which is containing that point . Form should look like this : $\left(\frac{\alpha x+\beta y+\lambda}{\sqrt{\alpha^{2}+\beta^{2}}}\right)^{2}= \frac{1}{√(\alpha^2 + \beta^2)}\left(\frac{\beta x-\alpha y+\mu}{\sqrt{\left(\alpha^{2}+\beta^{2}\right)}}\right)$

ProblemDestroyer
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  • With directrix $ax+by+c=0$ and focus $F=(h,k)$ the equation of the parabola is $$\frac{(ax+by+c)^2}{a^2+b^2}=(x-h)^2+(y-k)^2\tag1$$

    The axis is $bx-ay+d=0,$ where $d$ is determined by the fact that the focus is on the axis. $d=ak-bh.$

    The vertex then is determined by the intersection $\langle bx-ay+ak-bh,\frac{(ax+by+c)^2}{a^2+b^2}=(x-h)^2+(y-k)^2\rangle$ or $(x,y)=(-(abk+((-2b^2)-a^2)h+ac)/(2b^2+2a^2),((b^2+2a^2)k-abh-bc)/(2b^2+2a^2)).$

    The tangent at vertex goes through the vertex and is parallel to the directrix: $ax+by-(bk+ah-c)/2=0$

    $ax+by-(bk+ah-c)/2=M(bx-ay+ak-bh)^2\tag2$

     – Jan-Magnus Økland Apr 18 '22 at 14:35
  • Comparing coefficients $(1)=L(2)$

    $L = -2c-2kb-2ha, M = 1/(2c+2kb+2ha)$

    Making the tangent at vertex form of the equation $$ax+by-(bk+ah-c)/2=(bx-ay+ak-bh)^2/(2c+2kb+2ha)$$ and $(1)$ eqiuvalent, when $-2c-2kb-2ha\neq 0.$

     – Jan-Magnus Økland Apr 18 '22 at 14:36
  • Nice one @Jan-MagnusØkland – ProblemDestroyer Apr 18 '22 at 19:04

2 Answers2

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You can take the axis of the parabola to be the $(x, y) = (h, k) + t (\cos \theta, \sin(\theta) $, where $(h,k)$ is the focus. Then the directrix equation is

$ \cos(\theta) x + \sin(\theta) y = K $

where $ K = -a + h \cos(\theta) + k \sin(\theta) $

The equation of the parabola is

$ (x - h)^2 + (y - k)^2 = ( \cos(\theta) (x - h) + \sin(\theta) (y - k) +a )^2 $

Hosam Hajeer
  • 21,978
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Looking at the equations $$ u =\alpha x + \beta y + \lambda $$ $$ v = \beta x - \alpha y + \mu $$ one can see that the first line is mapped to the $u=0$ coordinate lines and the second to $v=0$. To ensure that this map preserves distance we need to normalize it: $$ u = \frac{1}{{\sqrt{\alpha^2+\beta^2}}} (\alpha x + \beta y + \lambda) \\ $$ $$ v = \frac{1}{{\sqrt{\alpha^2+\beta^2}}} (\beta x - \alpha y + \mu) $$ In the $u,v$ plane the equation for the parabola with the focal distance $f$ and the axis $u=0$ is: $$ v = \frac{u^2}{4f} $$ if we plug in the expressions for $u$ and $v$ we get $$ \frac{- \alpha y + \beta x + \mu}{\sqrt{\alpha^{2} + \beta^{2}}} - \frac{\left(\alpha x + \beta y + \lambda\right)^{2}}{4 f \left(\alpha^{2} + \beta^{2}\right)} = 0 $$ After getting rid of the denominators we get: $$ {4 f \sqrt{\alpha^{2} + \beta^{2}}} \left({- \alpha y + \beta x + \mu}\right) = {\left(\alpha x + \beta y + \lambda\right)^{2}} $$

blamocur
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