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I've been working through some past papers for an exam which I am due to be sitting tomorrow. In the Conic Sections paper from a couple of years ago, the following question came up:


The path of a comet can be modeled by a parabola with equation $y^2=2x$.

The Sun, at S, is at the focus of the parabola, and the Earth, at E, lies along the axis of symmetry of the parabola at a distance of 1 Astronomical Unit from the Sun.

diagram

What is the closest the comet gets to Earth?

Notes: The movement of the Sun and the Earth can be ignored. An Astronomical Unit (AU) is the average distance from the Earth to the Sun.

Source: http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2009/90639-exm-09.pdf


The parabola adopts the form $y^2=4ax$ so the focus is located at $(a,0)$. $4a=2$ so $a=\frac{1}{2}$. Therefore, S is located at $(\frac{1}{2},0)$ and E is located at $(\frac{3}{2},0)$.

Answers are available for this question, telling me that the shortest distance is $\sqrt{2}$ AU, however I am unable to formulate how this has been derived. Would you mind lending a hand?

Thank you.

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    Ah, my apologies. I was viewing $(x,0)$ as the focus. Thank you very much for your help! =)2011-11-23

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Any point on the parabola is of the form $(b^2/2,b)$. The square of the distance from Earth of this point is $s^2=(b^2-3/2)^2+b^2$. To minimize this, take the derivative with respect to $b$, set it to zero, and you will get the value of $b$. One root will be $b=0$, you can ignore that one as it is a local maximum.