Distance from Ellipse to Point Q (1 Viewer)

mreditor16 · May 4, 2015

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

)

thanks!!

InteGrand · May 4, 2015

mreditor16 said:
Help with this question would be greatly appreciated (by me and tonnes of my friends )

thanks!!

A general point on the ellipse is $P(x,y)$ , satisfying the condition $\frac{x^2}{4}+\frac{y^2}{1}=1$ .

The squared distance from P to (a, 0) is $s=(x-a)^2 + y^2$ .

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

InteGrand · May 4, 2015

mreditor16 said:
Help with this question would be greatly appreciated (by me and tonnes of my friends )

thanks!!

Also, just geometrically, the following should be clear: maximal distance is always $D_\text{max.}=a+2$ , and for $a\geq 2$ , $D_\text{min.}$ will always be $a-2$ .

mreditor16 · May 4, 2015

InteGrand said:
A general point on the ellipse is $P(x,y)$ , satisfying the condition $\frac{x^2}{4}+\frac{y^2}{1}=1$ .

The squared distance from P to (a, 0) is $s=(x-a)^2 + y^2$ .

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 · May 4, 2015

Also for $D_\text{min.}$ , we will always have $x\geq 0$ for the point on the ellipse that is closest to (a, 0).

InteGrand · May 4, 2015

mreditor16 said:
I have learnt neither of those methods. :/ I am a first year! :/

Using HSC methods, we can rearrange the condition $\frac{x^2}{4}+\frac{y^2}{1}=1$ 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.

mreditor16 · May 4, 2015

InteGrand said:
Also, just geometrically, the following should be clear: maximal distance is always $D_\text{max.}=a+2$ , and for $a\geq 2$ , $D_\text{min.}$ will always be $a-2$ .

Answers say :/

InteGrand · May 4, 2015

mreditor16 said:
Answers say :/

OK. (That doesn't contradict anything I said.)

ymcaec · May 4, 2015

mreditor16 said:
Help with this question would be greatly appreciated (by me and tonnes of my friends )

thanks!!

For $a\geq 2$ shortest distance is $\left | a-2 \right |$

Let a point on the ellipse to be $\left ( x_{1},y_{1} \right )$ .

Distance from $\left ( a,0 \right )$ to the point on ellipse is given by $\sqrt{(x_{1}-a)^2+y_{1}^2}$ .

We can use the equation of the ellipse to replace $y_{1}$ by $x_{1}$ . We get:

$d = \sqrt{x_{1}^2-2ax_{1}+a^2+1-\frac{x_{1}^2}{4}}\\d=\sqrt{(\frac{\sqrt{3}}{2}x_{1}-\frac{2}{\sqrt{3}}a)^2-\frac{1}{3}a^2+1}$

d is minimised when $x$ is chosen such that $(\frac{\sqrt{3}}{2}x_{1}-\frac{2}{\sqrt{3}}a)^2 = 0$

Thus, shortest distance is $d=\sqrt{-\frac{1}{3}a^2+1}$ .

When $a=\sqrt{3},$ $ d = 0$ which is not true, so we know that $d=\sqrt{-\frac{1}{3}a^2+1}$ is only true up to some $a$ .

Now consider the intersection of $\sqrt{-\frac{1}{3}a^2+1}$ and $\left | a-2 \right | \Rightarrow a = 1.5$ .

So for $a \geq 1.5$ we shall use $d = \left | a-2 \right |$ .

Thus $\left\{\begin{matrix}\sqrt{1-\frac{1}{3}a^2} & $for $0\leq a\leq \frac{3}{2}\\ \left | a-2 \right | & $for $ a > \frac{3}{2}\end{matrix}\right$

InteGrand · May 4, 2015

So the ellipse condition is that the point P on the (upper-half of the) ellipse has $y = \sqrt{1-\frac{x^2}{4}}$ .

The squared distance as a function of x is then $s(x) = (x-a)^2 + 1-\frac{x^2}{4}$ .

This can be differentiated to help find the minimum:

$s^\prime (x) = 2(x-a)-\frac{2x}{4}=2(x-a)-\frac{x}{2}=\frac{3}{2}x-2a$

$\Rightarrow s^\prime (x) = 0 \text{ when }\frac{3}{2}x-2a=0\Leftrightarrow x=\frac{4a}{3}$ .

So the minimal squared distance is $s_\text{min.}=\left(\frac{4a}{3} -a \right)^2 + 1 - \frac{\left(\frac{4a}{3} \right)^2}{4}=\frac{1}{3}(3-a)^2 = 1 - \frac{a^2}{3}$

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 $1-\frac{x^2}{4}\geq 0$ (since it's under a square root). Subsituting $x=\frac{4a}{3}$ yields $a\leq \frac{3}{2}$ in order for the top half of the answer to be feasible.

InteGrand · May 4, 2015

(For $a>\frac{3}{2}$ , the point on the ellipse we worked out as being closest, namely $(x_\text{min.},y_\text{min.}) =\left(\frac{4a}{3},\sqrt{1-\frac{\left(\frac{4a}{3}\right)^2}{4}}\right)$ , is no longer a real point, which is why we need the condition on a.)

mreditor16 · May 4, 2015

Oh wow, so tedious! :/ Thanks guys!!

InteGrand · May 4, 2015

To see why using $D_\text{min.}=|a-2|$ for $\frac{3}{2}<a\leq 2$ 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 $x_\text{vert}=\frac{4a}{3}>2$ as $a>\frac{3}{2}$ . 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 $2-a$ , which is the same as $|a-2|$ as required.

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

zhertec · May 4, 2015

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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Distance from Ellipse to Point Q (1 Viewer)

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