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Given 2 points. $M(-3; 2)$ and $N(2; 5)$

Find a point $P$ which is on Y axis, so that $distance(M, P) - distance(N, P)$ is maximal

From condition i assumed that P has coordinates $P(0; y)$

How to solve it?

Daniel
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  • Your question is not different from this one. – najayaz Oct 18 '15 at 14:11
  • @G-man I think they are different. According to solution given in problem # 1483390, P should be on the same line as MN and on the y-axis. This gives P= (0, 3.8). and d = PM – PN = 3.5 – 2.33 = 1.17. However, if I let P be at P’(0, –50(roughly)), then, through Geogegra, I have P’N – P’M = 53.77 – 50.83 which is way larger than d. – Mick Oct 18 '15 at 14:53
  • @Mick if we assume an equation for Y axis, so x=0, maybe that solution would work. – Daniel Oct 18 '15 at 15:31
  • x = 0 is a must because P is a point on the y-axis. – Mick Oct 18 '15 at 16:00
  • @Mick I just glanced at this question and thought the concept involved was the same. – najayaz Oct 18 '15 at 16:08
  • @G-man No problem. – Mick Oct 19 '15 at 04:43

3 Answers3

1

If sum of distances $ MP+ PN$ is constant =$d_1$ , an ellipse can be defined tangent to y-axis only for a particular $d_1$.

Similarly,

If difference of distances $ MP-PN $ is constant =$d_2$ , a hyperbola can be defined tangent to y-axis only for a particular $d_2$.

Point of tangency / contact has a zero x- coordinate, non-zero y-coordinate.

Numerical results:

Distance difference $ =d_2 \approx 3.18 $, point of tangency $ \approx (0,10).$

Doppler

Narasimham
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  • calculation with a brute-force algorithm gives me (0, 11.00) it's close to your answer. But i don't fully understand your solution, can you draw for me to undersand better? Thanks. – Daniel Oct 18 '15 at 15:27
  • What was the distance difference in your construction? – Narasimham Oct 18 '15 at 15:33
  • dM = 9.487 dN = 6.325 delta = 3.162 y = 11.000 – Daniel Oct 18 '15 at 15:38
  • Seems alright. I read from graph y=10 by eye estimation as you can see it is narrow , 11.0 would be more accurate. – Narasimham Oct 18 '15 at 17:36
  • btw i dont understand this image. Can you write step to step solution with images? Of course if you want. – Daniel Oct 18 '15 at 17:58
  • After knowing what to expect, I graphed it on Mma. ContourPlot[ {x==0,Sqrt[(x+3)^2 + (y-2)^2] - Sqrt[(x-2)^2 + (y-5)^2] ==3.18}, {x,-0,10},y,0,15},GridLines->Automatic,AspectRatio->Automatic,ContourStyle->{Thick,Red}]. It avoids cumbersome simplifications. – Narasimham Oct 18 '15 at 18:07
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You are correct in assuming the point $P$ will have coordinates $(0, y)$.

Edit: My original $d_M$ and $d_N$ were incorrect. They have now been corrected. The distance from $M$ to $P$, here denoted $d_M =\sqrt{9+(y-2)^2}$, and the distance from $N$ to $P$ can be written $d_N =\sqrt{4+(y-5)^2}$. $d_M - d_N$ can now be written as a function of $y$, and we can find the maximum value by deriving. $\frac{d}{dy} \sqrt{9+(y-2)^2} -\sqrt{4+(y-5)^2} =$ $\frac{y-2}{\sqrt{9+(y-2)^2}} - \frac{y-5}{\sqrt{4+(y-5)^2}} = 0$ $\frac{(y-2)^2}{9+(y-2)^2} = \frac{(y-5)^2}{4+(y-5)^2}$ $\frac{(y-2)^2+9-9}{(y-2)^2+9} = \frac{(y-5)^2+4-4}{(y-5)^2+4}$ $1-\frac{9}{(y-2)^2+9} = 1-\frac{4}{(y-5)^2+4}$ $\frac{9}{(y-2)^2+9} = \frac{4}{(y-5)^2+4}$ $9(4+(y-5)^2) = 4(9+(y-2)^2)$ $9(y-5)^2 = 4(y-2)^2$ $9y^2 - 90y + 225 = 4y^2 - 16y +16$ $5y^2 - 74y + 209 = 0$ $y = \{11, \frac{38}{10}\}$ Plugging these values back into $d_M - d_N$, we find $y(11) = \sqrt{10}$ and $y(38/10) = \sqrt{34}/5$, so the answer is $P = (0, 11)$
Rob
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This is NOT a solution but is an actual plot of the situation.

enter image description here

Some of the data have been given in my comment above.

When we consider P is on the positive y-axis, say at P’’(0, 10(roughly)), D (= P’’M – P’’N = MX = 3.16 (roughly) as @Narasimham showed) has attained its maximum.

Mick
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