What is the Equation of the Locus of Points in the Hyperbola or Ellipse Problem?

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1. Homework Statement

The following question is posed within a section of my A level maths book titled "The Hyperbola"

A set of points is such that each point is three times as far from the y-axis as it is from the point (4,0). Find the equation of the locus of P and sketch the locus

2. Homework Equations 3. The Attempt at a Solution

If P is a point (x,y) on the locus, and N is the intersection on the y-axis of the line through P, parallel to the x axis, and S is the point (4,0) then
<br /> PN = 3PS\\<br /> <br /> PN^2 = 9PS^2\\<br /> x^2 = 9((x-4)^2+y^2)\\<br /> \frac{8}{9}x^2+y^2-8x+16=0\\<br />
Which I believe is an elipse, but my book indicates that it is a hyperbola with it's answer of
<br /> 8x^2-y^2+8x-16=0<br />

Is my book wrong?
 
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Appleton said:
1. Homework Statement

The following question is posed within a section of my A level maths book titled "The Hyperbola"

A set of points is such that each point is three times as far from the y-axis as it is from the point (4,0). Find the equation of the locus of P and sketch the locus

2. Homework Equations 3. The Attempt at a Solution

If P is a point (x,y) on the locus, and N is the intersection on the y-axis of the line through P, parallel to the x axis, and S is the point (4,0) then
<br /> PN = 3PS\\<br /> <br /> PN^2 = 9PS^2\\<br /> x^2 = 9((x-4)^2+y^2)\\<br /> \frac{8}{9}x^2+y^2-8x+16=0\\<br />
Which I believe is an elipse, but my book indicates that it is a hyperbola with it's answer of
<br /> 8x^2-y^2+8x-16=0<br />

Is my book wrong?

Yes, YOU are right. You can even plot the curve in some package such as Maple to see what is happening.

You can even argue intuitively that the curve must be bounded in the plane, because if you could take ##x \to \infty ## very large (and ##y## moderate) on the curve you would have have (approximately) ##x \approx 3 (x-4)##, so ##x \approx 6##, contradicting the condition that ##x \to \infty## is very large.
 
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Ray Vickson said:
Yes, YOU are right. You can even plot the curve in some package such as Maple to see what is happening.

You can even argue intuitively that the curve must be bounded in the plane, because if you could take ##x \to \infty ## very large (and ##y## moderate) on the curve you would have have (approximately) ##x \approx 3 (x-4)##, so ##x \approx 6##, contradicting the condition that ##x \to \infty## is very large.
Thanks, it's encouraging to know that writers of maths books fall prey to the same kind of mistakes that I do, if a little less frequently.
 
Appleton said:
Thanks, it's encouraging to know that writers of maths books fall prey to the same kind of mistakes that I do, if a little less frequently.
Can happen. The hyperbola they give as result is the solution of the exercise "A set of points is such that each point is three times as far from the y axis point (4,0) as it is from the point (4,0) y axis."
 
Appleton said:
Thanks, it's encouraging to know that writers of maths books fall prey to the same kind of mistakes that I do, if a little less frequently.

It would have been a hyperbola if it had said "... is 3 times as far from the point (4,0) as from the y axis".
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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