MHB What is the ratio of PA to PB on a semicircle with points A and B on the x-axis?

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The discussion centers on finding the ratio of distances PA to PB on a semicircle defined by the equation y = sqrt{1 - x^2}, with points A and B located on the x-axis at coordinates (a, 0) and (1/a, 0), respectively. The arbitrary point P is located on the semicircle in quadrant 2, with coordinates (x, sqrt{1 - x^2}). The distances from P to A and P to B are calculated using the distance formula, resulting in d(PA) = sqrt{a^2 - 2ax + 1} and d(PB) = (sqrt{a^2 - 2ax + 1})/a. The ratio PA/PB simplifies to a, confirming that PA is proportional to the value of a. The conclusion demonstrates the relationship between the distances and the parameter a.
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Semicircle y = sqrt{1 - x^2} is given as a graph on the xy-plane. Points A and B lie on the line y = 0. The x-coordinates of the points A and B are a and 1/a, respectively. Point P is an arbitrary point on the graph of y in quadrant 2 connecting to points A and B. Show that PA/PB = a.

Note: Assume that 0 < a < 1

Obviously, I need to find the distance from P to A and the distance from P to B.

Point A = (a, 0)

Point B = (1/a, 0)

I do not know the coordinates of point P.

I am stuck here.
 
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the coordinates of point P are $(x, \sqrt{1-x^2})$ ...
 
Thank you for providing the coordinates of point P.

Let d(PA) = distance from P to A.

After plugging into the distance formula for points on the xy-plane, and simplifying the radicand, I found d(PA) to be sqrt{a^2 - 2ax + 1 }.

Let d(PB) = distance from P to B.

Applying the same steps as before, I found d(PB) to be
(sqrt{a^2 -2ax + 1})/a.

PA/PB = a

[sqrt{a^2 - 2ax + 1}]/[sqrt{a^2 - 2ax + 1 }]/a = a

sqrt{a^2 - 2ax + 1 } • a/sqrt{a^2 - 2ax + 1 } = a

a = a

Done!
 
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