something esoteric? (finding locus of points by 2nd method)

sujoy

Junior Member
Joined
Apr 30, 2005
Messages
110
There is a problem that I have solved by one method. However, another method does not solve it.

Question: The sum of the distances of a movable point P from the two fixed points
(c, 0) and (-c, 0) is a constant value of 2a. Prove that the the equation to the locus of P is given by:

. . .\(\displaystyle \L \frac{x^2}{a^2}\, +\, \frac{y^2}{b^2}\, =\,1\)

...where a<sup>2</sup> - b<sup>2</sup> = c<sup>2</sup>.
[Nothing more is given; the above is the full statement of the question.]

I have assumed the point P to have coordinates (x, y).

By the Distance Formula we get:

. . .Equation 1:

. . .\(\displaystyle \L \sqrt{(x\,-\,c)^2\, +\,y^2}\, +\, \sqrt{(x+c)^2 + y^2}\, =\,2a\)

Then:

. . .\(\displaystyle \L \frac{1}{\sqrt{(x\,-\,c)^2\,+\,y^2}\,+ \,sqrt{(x\,+\,c)^2\,+\,y^2}}\, =\,\frac{1}{2a}\)

Rationaliting the denominator, we get:

sqrt[(x-c)^2 +y^2] - sqrt[(x+c)^2 + y^2]
=>---------------------------------------------------------- = 1/(2a)**********[important1]
-2cx -2cx

sqrt[(x-c)^2 +y^2] - sqrt[(x+c)^2 + y^2] = [-2cx]/a _________________(a2eq)

adding 1eq&2eq we get ; 2sqrt[(x-c)^2 +y^2] = 2(a^2 - cx)/a
now proceeding accordingly we get the result,

However
sqrt[(x+c)^2 +y^2] - sqrt[(x-c)^2 + y^2]
=>---------------------------------------------------------- = 1/(2a)**********[important2]
-2cx -2cx
does not give the result.

Q1) Why could this be, or am I going wrong?

Q2) If I am correct, then what does this dilemma point to?

Regards,
Sujoy
 
I can't translate what you have written. Can you improve the notation - maybe LaTeX?

Generally, the method should not interfere with the result. The most likely suspect on the success of one and the failure of another is user error.
 
Sujoy: I have edited your post, as best I could, to clarify what you appear to be saying. But I couldn't make heads or tails of what comes after "Rationalizing the denominator", so you'll have to clarify the rest.

Thank you.

Eliz.
 
It appears they want you to derive the equation of an ellipse knowing that the sum of the distances from some point on the ellipse to the foci is 2a.

This should be in most any algebra text which covers conic sections.

You shouldn't need the text though.

Transpose eqaution 1 to the right side and square:

\(\displaystyle \L\\\sqrt{(x+c)^{2}+y^{2}}=2a-\sqrt{(x-c)^{2}+y^{2}}\)

\(\displaystyle \L\\=(x+c)^{2}+y^{2}=4a^{2}-4a\sqrt{(x-c)^{2}+y^{2}}+(x-c)^{2}+y^{2}\)

Simplify:

\(\displaystyle \L\\\sqrt{(x-c)^{2}+y^{2}}=a-\frac{c}{a}x\)

Square again and simplify:

\(\displaystyle \L\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-c^{2}}=1\)

Now, because of \(\displaystyle a=\sqrt{b^{2}+c^{2}}\), we can write it as:

\(\displaystyle \L\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Does this help?.
 
please could anyone of you give me the code for formatting like for say LaTeX etc
or refer me on a website to this effect---I understand the difficulty

on our problem :-
2sqrt[(x-c)^2 +y^2] = 2(a^2 - cx)/a ___________________________step1
=> sqrt [(x - c )^2 + y^2] = (a^2 - cx)/a
=> (x - c)^2 + y^2 = ( a^2 - cx)^2/a^2
=> x^2 +c^2 +y^2 - 2xc = [a^4 + c^2x^2 - 2a^2cx]/a^2
..................now a^2 - b^2 = c^2
=> (ax)^2 + (ac)^2 + (ay)^2 - 2xca^2
= a^4 + (xa)^2 - (bx)^2 - 2xa^2
Cancelling :
(ac)^2 + (ay)^2 = a^4 - (bx)^2
now a^2 - b^2 = c^2 again
a^4 -(ab)^2 +(ay)^2 = a^4 - (bx)^2 again cancelling:
(bx)^2 + (ay)^2 = (ab)^2. dividing this by (ab)^2
(x/a)^2 + (y/b)^2 = 1 which proves the fact as asked by the problem
However if we had written a step as this [important 2] :
sqrt[(x+c)^2 +y^2] - sqrt[(x-c)^2 + y^2]
=>---------------------------------------------------------- = 1/(2a]
-2cx -2cx
then instead of getting our "step1" we would have got
2sqrt[(x+c)^2 +y^2] = 2(a^2 - cx)/a ----------could we prove the problem from here also ....... here we go as :
sqrt [(x +c )^2 + y^2] = (a^2 - cx)/a
=> (x + c)^2 + y^2 = ( a^2 - cx)^2/a^2
=> x^2 +c^2 +y^2 + 2xc = [a^4 + c^2x^2 - 2a^2cx]/a^2
..................now a^2 - b^2 = c^2
=> (ax)^2 + (ac)^2 + (ay)^2 + 2xca^2
= a^4 + (xa)^2 - (bx)^2 - 2xa^2
Cancelling :
(ac)^2 + (ay)^2 = a^4 - (bx)^2 -4xca^2
now a^2 - b^2 = c^2 again
a^4 -(ab)^2 +(ay)^2 = a^4 - (bx)^2 -4xca^2 again cancelling:
(bx)^2 + (ay)^2 = (ab)^2 -4xca^2 . dividing this by (ab)^2
(x/a)^2 + (y/b)^2 = 1 -4xc/b^2 which does not proves the fact why?
thank you
Sujoy


[/code]
 
Was I on the right track with what you were trying to do?.

For LaTex, click on the quote at the upper right hand corner of my post to see the code used. Do a google search and many sites will pop up regarding LaTex.
 
Galactus you were perfectly correct with your approach to the problem , but if you note this part given below you might find that this way the solution is not reached-------this is what I was asking is about the reason[if this phenomenon is correct]
take care.
regards
Sujoy

However if we had written a step as this [important 2] :
sqrt[(x+c)^2 +y^2] - sqrt[(x-c)^2 + y^2]
=>---------------------------------------------------------- = 1/(2a]
-2cx -2cx
then instead of getting our "step1" we would have got
2sqrt[(x+c)^2 +y^2] = 2(a^2 - cx)/a ----------could we prove the problem from here also ....... here we go as :
sqrt [(x +c )^2 + y^2] = (a^2 - cx)/a
=> (x + c)^2 + y^2 = ( a^2 - cx)^2/a^2
=> x^2 +c^2 +y^2 + 2xc = [a^4 + c^2x^2 - 2a^2cx]/a^2
..................now a^2 - b^2 = c^2
=> (ax)^2 + (ac)^2 + (ay)^2 + 2xca^2
= a^4 + (xa)^2 - (bx)^2 - 2xa^2
Cancelling :
(ac)^2 + (ay)^2 = a^4 - (bx)^2 -4xca^2
now a^2 - b^2 = c^2 again
a^4 -(ab)^2 +(ay)^2 = a^4 - (bx)^2 -4xca^2 again cancelling:
(bx)^2 + (ay)^2 = (ab)^2 -4xca^2 . dividing this by (ab)^2
(x/a)^2 + (y/b)^2 = 1 -4xc/b^2 which does not proves the fact why?
thank you
Sujoy
 
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