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I solved the equation up till:

$$ (x-y)^2 = 2a(x+y-\frac{a}{2}) $$

I'm not sure how to proceed from here. Correct answer is $ a\sqrt{2} $, shouldn't it be $ 2a $?

Bach
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Guruguru
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    Related: In this solution, I show that $$\sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}}=1$$ is a parabola with focus $F = \dfrac{ab}{c^2}(b,a)$ (where $c^2=a^2+b^2$) and directrix $a x + b y = 0$. Since the semi-latus rectum of a parabola is the distance from focus to directrix, we have $$\text{latus rectum} = 2; \frac{a(ab^2/c^2)+b(a^2b/c^2)}{\sqrt{a^2+b^2}} =\frac{4a^2b^2}{c^3}$$ In your case, $a=b$, so the answer $4a^4/(2a^2)^{3/2} = a \sqrt{2}$. – Blue May 03 '18 at 01:28
  • @Blue Good stuff. Thanks mate – Guruguru May 03 '18 at 03:55

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It's not $2a$ because the change of variables $u=x-y$, $v=x+y$ is not orthogonal, and so it changes lengths and distorts objects. To see the correct values, we can make the following change of variables: $$u=\frac{x-y}{\sqrt{2}}, \quad v=\frac{x+y}{\sqrt{2}},$$ which is effectively a rotation of the coordinate plane by $45^\circ$. Then the equation of the parabola becomes $$u^2=\sqrt{2}a\left(v-\frac{a}{2\sqrt{2}}\right),$$ and we can see that the latus rectum is $\sqrt{2}a$.

zipirovich
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