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.


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.

  • $\begingroup$ So, you know how to get the distance from $(3/2,0)$ to a point on the parabola, no? The square of the distance is a quadratic itself, and I presume you know how to get its minimum... $\endgroup$ – J. M. is a poor mathematician Nov 23 '11 at 4:55
  • $\begingroup$ The marking schedule gives $d^2=(1.5-x)^2+y^2$ as the equation for the distance from $(\frac{3}{2},0)$ to the parabola. This is the step that I cannot visualize. Please note that I am a 5th former who has been thrown into a 7th form class last minute. I'm sure that I am just missing something pathetically simple. Thanks. $\endgroup$ – Charles Salmon Nov 23 '11 at 5:08
  • $\begingroup$ ...and you know that $y^2=2x$, right? Now your expression for $d^2$ is a nice quadratic in $x$ whose minimum you can easily find... $\endgroup$ – J. M. is a poor mathematician Nov 23 '11 at 5:12
  • $\begingroup$ Yes. But I don't know how the equation for the distance from $(\frac{3}{2},0)$ came about. $\endgroup$ – Charles Salmon Nov 23 '11 at 5:13
  • $\begingroup$ It's an application of the Pythagorean theorem. Take a right triangle whose hypotenuse is the segment joining $(3/2,0)$ and the parabola point, and whose legs are parallel to the coordinate axes. $\endgroup$ – J. M. is a poor mathematician Nov 23 '11 at 5:20

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.


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