Distance of a point on the ellipse

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dr.trovacek

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This is a problem from the chapter derivates – fourth grade high school

Problem: Determine the point on the ellipse that is the furthest from the far end of the minor axis. What is that distance?

So we have to find the point on the ellipse that has the largest distance from the co-vertex point \((0,b) \) or \((0,-b) \).

I know how to take a derivative to determine a maximum of an expression. So I first need to express the x or y from the equation of the ellipse and construct an expression (function) of which I could find the maximum.

In the textbook they say there are two cases:
  1. when the minor (vertical) axis is greater than the major (horizontal), then the distance equals \( d =2b \)
  2. when the minor (vertical) axis is smaller than the major (horizontal), the distance equals \( d^2 = x^ 2 + (b + y)^2 \)


So I think I understand this conceptually:
  1. When the vertical axis is longer than the horizontal, then the largest distance is at the far end point of the axis and the distance is 2b.
  2. When the vertical axis is longer – lets say we look at a far end point \( (0,b) \) – then the largest distance is in two points (symmetrical with respect to y axis) located somewhere around the lower far point end, and this is calculated via coordinates – the second term on the right side of the distance equation is \( (b+y)^2 \) because we are looking for a point under the x axis.

There is a condition given for each of the latter cases:
  1. \( a ^ 2≤2b ^ 2 \)
  2. \( a^2>2b^2 \)

So the \( a^2 = 2b \) is the border between the two cases.

I was just wondering if someone could explain how to reach the conclusion of the condition? Is there a geometrical explanation using the coordinates? It is not clear to me how to conclude that the change between cases is happening exactly at \( a^2 = 2b ^ 2 \) .
 
I'm not sure I understand. I would not start with those cases; I would obtain them as the result of the work.

I'd just write the expression for the distance (as you have under 2) and maximize that under the condition that (x,y) is on the ellipse. Then I'd expect to find that the local maximum only exists under the condition they end up saying, and otherwise the opposite end of the minor axis (and endpoint of the domain) is the maximum. That would be "case 1", but it's a part of the answer, not part of the work, as I'd do it. No preliminary geometrical work should be needed.

What exactly does the textbook actually show? I'd like to see it.
 
I can't take a picture because the solution is not in English, but I used Mathpix Snipping Tool to get all the equations and translated the text parts.

Form the ellipse equation \( b ^ { 2 } x ^ { 2 } + a ^ { 2 } y ^ { 2 } = a ^ { 2 } b ^ { 2 }\) follows \( x ^ { 2 } = \frac { a ^ { 2 } b ^ { 2 } - a ^ { 2 } y ^ { 2 } } { b ^ { 2 } } =\frac { a ^ { 2 } \left( b ^ { 2 } - y ^ { 2 } \right) } { b ^ { 2 } } \)
If \(a ^ { 2 } \leqslant 2 b ^ { 2 } \) that is \( a \leqslant b \sqrt { 2 } \) then \( d = 2 b \)

Let say that \( a > b \sqrt { 2 } \). Then \( d ^ { 2 } = x ^ { 2 } + ( b + y ) ^ { 2 } = \sqrt { \frac { a ^ { 2 } } { b ^ { 2 } } \left( b ^ { 2 } - y ^ { 2 } \right) + ( b + y ) ^ { 2 } } =\sqrt { a ^ { 2 } - \frac { a ^ { 2 } y ^ { 2 } } { b ^ { 2 } } + b ^ { 2 } + 2 y b + y ^ { 2 } } \)

If we take the derivative of a function \( d ( y ) = \sqrt { a ^ { 2 } - \frac { a ^ { 2 } y ^ { 2 } } { b ^ { 2 } } + b ^ { 2 } + 2 y b + y ^ { 2 } } \) we get \( \frac { - \frac { 2 a ^ { 2 } y } { b ^ { 2 } } + 2 b + 2 y } { 2 \sqrt { a ^ { 2 } - \frac { a ^ { 2 } y ^ { 2 } } { b ^ { 2 } } + b ^ { 2 } + 2 y b + y ^ { 2 } } } \)

We look when this is equal to zero \( - \frac { 2 a ^ { 2 } y } { b ^ { 2 } } + 2 b + 2 y =0 \Longrightarrow - a ^ { 2 } y + y b ^ { 2 } = - b ^ { 3 } \Longrightarrow y \left( a ^ { 2 } - b ^ { 2 } \right) = b ^ { 3 } \Longrightarrow y = \frac { b ^ { 3 } } { a ^ { 2 } - b ^ { 2 } } \\

x= \sqrt { \frac { a ^ { 2 } b ^ { 2 } - a ^ { 2 } \frac { b ^ { 6 } } { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } } { b ^ { 2 } } } = \sqrt { a ^ { 2 } - \frac { a ^ { 2 } b ^ { 4 } } { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } } = \frac { a ^ { 2 } \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } - a ^ { 2 } b ^ { 4 } } { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } = \frac { a ^ { 2 } \left[ \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } - b ^ { 4 } \right] } { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } = \frac { a ^ { 2 } \left[ \left( a ^ { 2 } - b ^ { 2 } - b ^ { 2 } \right) \left( a ^ { 2 } - b ^ { 2 } + b ^ { 2 } \right) \right] } { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } = \frac { a ^ { 4 } \left( a ^ { 2 } - 2 b ^ { 2 } \right) } { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } \)

The coordinates of the point T that we are looking for are \( T \left( \frac { a ^ { 2 } } { a ^ { 2 } - b ^ { 2 } } \sqrt { a ^ { 2 } - 2 b ^ { 2 } } , - \frac { b ^ { 3 } } { a ^ { 2 } - b ^ { 2 } } \right) \)



This is the entire solution given in the textbook. There is no explanation how they get to the conditions .
 
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Thanks.

I got the same answer, by more or less the same method, and I didn't make any assumptions about [MATH]a^2>2b^2[/MATH] and the like at the start. I just got to the end, and saw (from the radical in the expression for x) that this solution is only valid when [MATH]a^2>2b^2[/MATH]. When this is not true, the maximum has to be at one of the endpoints, so I would check them. I don't see how you would know ahead of time that the two cases would be what they are, and I don't think you need to. I would ignore it; I don't think there's anything important that you're missing.
 
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