Distance from Ellipse to Point Q (1 Viewer)

mreditor16

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Help with this question would be greatly appreciated (by me and tonnes of my friends :D)

thanks!! :)
 
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InteGrand

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Help with this question would be greatly appreciated (by me and tonnes of my friends :D)

thanks!! :)
A general point on the ellipse is , satisfying the condition .

The squared distance from P to (a, 0) is .

Now either you can either use Lagrange multipliers or rearrange the given condition to turn this into a one-variable optimisation problem.
 

InteGrand

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Help with this question would be greatly appreciated (by me and tonnes of my friends :D)

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Also, just geometrically, the following should be clear: maximal distance is always , and for , will always be .
 

mreditor16

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A general point on the ellipse is , satisfying the condition .

The squared distance from P to (a, 0) is .

Now either you can either use Lagrange multipliers or rearrange the given condition to turn this into a one-variable optimisation problem.
I have learnt neither of those methods. :/ I am a first year! :/
 

InteGrand

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Also for , we will always have for the point on the ellipse that is closest to (a, 0).
 

InteGrand

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I have learnt neither of those methods. :/ I am a first year! :/
Using HSC methods, we can rearrange the condition to get y in terms of x.

We can then plug this into our squared-distance function to get it as solely a function of x, which we can then seek to optimise using methods of single variable calculus.

Also, we can just let y ≥ 0.
 

ymcaec

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Help with this question would be greatly appreciated (by me and tonnes of my friends :D)

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For shortest distance is

Let a point on the ellipse to be .

Distance from to the point on ellipse is given by .

We can use the equation of the ellipse to replace by . We get:



d is minimised when is chosen such that

Thus, shortest distance is .

When which is not true, so we know that is only true up to some .

Now consider the intersection of and .

So for we shall use .

Thus
 

InteGrand

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So the ellipse condition is that the point P on the (upper-half of the) ellipse has .

The squared distance as a function of x is then .

This can be differentiated to help find the minimum:



.

So the minimal squared distance is

From simplifying this, you get the top part of the answer you posted.

To see the reason for the split in answers for different values in a, note that (since it's under a square root). Subsituting yields in order for the top half of the answer to be feasible.
 
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InteGrand

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(For , the point on the ellipse we worked out as being closest, namely , is no longer a real point, which is why we need the condition on a.)
 

InteGrand

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To see why using for is optimal, note that the squared distance function represents a concave up parabola. Therefore, it is minimised at its vertex, and as we saw before, this globally minimising x-value is as . So the parabola is minimised at a point where x > 2, so the closest we can get to this minimal value by taking a point on the ellipse is to make x on the ellipse closest to this value greater than 2, which is clearly by taking x = 2, and this point (the point (2,0)) on the ellipse has distance to the point (a, 0) of , which is the same as as required.

(And we noted earlier that the optimal distance for a > 2 was , which is equal to , so in either case is optimal.)
 
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zhertec

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On an unrelated note, I feel that integrand was like a state ranker for 3u/4u and is now doing like actuarial studies, am I correct in any sense? lol
 

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