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The problem is,

Find the focus, equation of directrix, vertex, length of latus rectum of the parabola given by, $$\left(\alpha x+\beta y+\gamma\right)^2=Ax+By+C$$

I am stuck with the problem for quite sometime but still don't know how to begin. Can anyone help?

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  • $\begingroup$ Is "rectum" seriously a mathematical word? Why am I just now finding out about this?! $\endgroup$ – Daniel W. Farlow Apr 3 '15 at 5:09
  • $\begingroup$ @crash yes it is, you'd know if you had taken up conics $\endgroup$ – G-man Apr 3 '15 at 5:18
  • $\begingroup$ @G-man I'd gladly have taken up the conics rather than take anything up the rectum. :D $\endgroup$ – Daniel W. Farlow Apr 3 '15 at 5:20
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    $\begingroup$ I'd have upvoted that comment but the tooltip saying "this comment adds something useful to the post" stops me from doing that. $\endgroup$ – G-man Apr 3 '15 at 5:26
  • $\begingroup$ @user170039 , this is problem is for the most part, mechanical and the results are quite complicated, especially so for the focus. Why do you want to do this problem? $\endgroup$ – G-man Apr 5 '15 at 15:39
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First simplify the parabola equation to a general $2^{nd}$ degree equation: $$(\alpha x+\beta y)^2+(2\beta\gamma - B)y+(2\alpha\gamma-A)x+\gamma^2-C=0$$

Now let $2g=2\alpha\gamma-A$,$2f=2\beta\gamma - B$ and $d=\gamma^2-C$ and then substitute in the equation: $$(\alpha x+\beta y)^2=-2fy-2gx-d$$ Let $\exists\lambda\in\mathbb R$ such that : $$(\alpha x+\beta y+\lambda)^2=2(\alpha\lambda-g)x+2(\beta\lambda-f)y+\lambda^2-d$$ Note that the equation of any parabola can be written in the form: $$(distance\;from\;axis)^2=(length\;of\;latus\;rectum)\times(distance\;from\;tangent\;at\;vertex)$$ For the lines on the LHS and RHS to be perpendicular to each other, $$\frac \alpha\beta\times\frac{\alpha\lambda-g}{\beta\lambda-f}=-1\implies \lambda=\frac{\alpha g+\beta f}{\alpha^2+\beta^2}$$ Now we rewrite our equation as: $$\left( \frac{\alpha x+\beta y+\lambda}{\sqrt{\alpha^2+\beta^2}} \right)^2=\frac{2\sqrt{(\alpha\lambda-g)^2+(\beta\lambda-f)^2}}{\alpha^2+\beta^2}\times\frac{2(\alpha\lambda-g)x+2(\beta\lambda-f)y+\lambda^2-d}{2\sqrt{(\alpha\lambda-g)^2+(\beta\lambda-f)^2}}$$

From here we can make several conclusions:

The equation of the axis is:$$\alpha x+\beta y +\lambda=0\tag{i}$$

The equation of tangent at vertex is:$$2(\alpha\lambda-g)x+2(\beta\lambda-f)y+\lambda^2-d=0\tag{ii}$$ From the intersection of above 2 lines, the vertex coordinates are:$$\left( \frac{2f\lambda-\beta\lambda^2-\beta d}{2(\beta g- \alpha f)}\;,\;\frac{2g\lambda-\alpha\lambda^2-\alpha d}{-2(\beta g- \alpha f)} \right)\tag{iii}$$ The lentgh of latus rectum is $$L=\frac{2\sqrt{(\alpha\lambda-g)^2+(\beta\lambda-f)^2}}{\alpha^2+\beta^2}\tag{iv}$$

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