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This is an extension my earlier questions here and here on parabolas.

Find the length of the Latus Rectum of the General Parabola $$(Ax+Cy)^2+Dx+Ey+F=0$$

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Hypergeometricx
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  • To help you, we need to know what your background in this question is, and what you’ve tried for solving it. – Lubin Dec 09 '16 at 15:48
  • @Lubin - Thanks for reading. I've just posted an answer. Have worked this out after the two earlier questions, and another question by someone else on the LR of a parametric parabola. – Hypergeometricx Dec 09 '16 at 15:52

1 Answers1

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$$(Ax+Cy)^2+Dx+Ey+F=0\qquad\cdots (1)$$ Put $u=Ax+Cy, v=Cx-Ay$ and solving for $x,y$ and putting $K=A^2+C^2$ gives $$y=\frac {Cu-Av}{K}; \qquad x=\frac {Au+Cv}{K}\qquad\cdots (2)$$ From $(1),(2)$,

$$\begin{align} u^2+D\left(\frac {Au+Cv}{K}\right)+E\left(\frac{Cu-Av}{K}\right)+F&=0\\ \left(u+\frac {AD+CE}{2K}\right)^2&=\frac{AE-CD}{K}\bigg\lbrace v+\frac{K}{AE-CD}\left[\left(\frac{AD+CE}{2K}\right)^2-F\right]\bigg\rbrace\\ \text{Dividing by }\sqrt{K}\cdot \sqrt{K} \text{ to normalize }u,v \hspace{1cm} \\ \text{and putting }M=AE-CD,\hspace{1cm}\\ {\underbrace{\left(\frac u{\sqrt{K}}+\frac {AD+CE}{2K\sqrt{K}}\right)}_U}^2&=\underbrace{\frac{M}{K\sqrt{K}}}_{4\alpha}\underbrace{\bigg\lbrace \frac v{\sqrt{K}}+\frac{K}{M\sqrt{K}}\left[\left(\frac{AD+CE}{2K}\right)^2-F\right]\bigg\rbrace}_V\\ U^2&=4\alpha V\\ \text{Length of Latus Rectum } = |4\alpha |&=\bigg|\frac M{\;\;K^{\frac 32}}\bigg|=\color{red}{\frac {\big|AE-CD\big|}{\;\;\big(A^2+C^2\big)^{\frac 32}}} \end{align}$$

Hypergeometricx
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  • I thought the latus rectum was a chord. Are you calculating its length? – Lubin Dec 09 '16 at 15:56
  • @Lubin - Yes, that's right. Edited accordingly. Thanks. – Hypergeometricx Dec 09 '16 at 16:06
  • Something’s off. Set $A=\sqrt a$, $E=-1$ and the other coefficients to $0$ for the equation $y=ax^2$. Your formula produces $\sqrt a/a$, but the correct result is $1/a$. Compared to this, your result is missing a factor of $\sqrt{A^2+C^2}$ in the denominator. – amd Dec 10 '16 at 03:46
  • Hmmm how could this arise? The method substituting $u,v$ is effectively rotating the parabola, and completing the square etc is effectively translating it to the origin, and the steps seem quite procedural. Why does it give a different answer? – Hypergeometricx Dec 10 '16 at 04:29
  • @amd - OK, sorted out. – Hypergeometricx Dec 10 '16 at 07:56