(a) Note that (-1,0) lies on the curve x? + y² = 1. For t e R, denote by le the line with slope t that passes through (-1,0). This line intersects the curve x² + y? = 1 at (-1,0) as well as at one additional point (x(t), y(t)). In fact, since every point on the curve x² + y? = 1 other than (-1,0) is realized as the nontrivial intersection point of the line li for precisely one value of t, we obtain the parametrization E R² 2? + y? = 1} = {(#(t), y(t)) | t e R}U{(-1,0)}. Compute the values of x(t) and y(t) to obtain a parametrization of the real points lying on the curve x² + y² = 1.
(a) Note that (-1,0) lies on the curve x? + y² = 1. For t e R, denote by le the line with slope t that passes through (-1,0). This line intersects the curve x² + y? = 1 at (-1,0) as well as at one additional point (x(t), y(t)). In fact, since every point on the curve x² + y? = 1 other than (-1,0) is realized as the nontrivial intersection point of the line li for precisely one value of t, we obtain the parametrization E R² 2? + y? = 1} = {(#(t), y(t)) | t e R}U{(-1,0)}. Compute the values of x(t) and y(t) to obtain a parametrization of the real points lying on the curve x² + y² = 1.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:(a) Note that (-1,0) lies on the curve x? + y? = 1. For t e R, denote by le the
line with slope t that passes through (-1, 0). This line intersects the curve
x² + y?
1 at (–1,0) as well as at one additional point (x(t), y(t)). In fact,
1 other than (-1,0) is realized as the
since every point on the curve x² + y?
nontrivial intersection point of the line l4 for precisely one value of t, we obtain
the parametrization
{(x, y) € R² | x² + y² = 1} = {(x(t), y(t)) |t € R}U{(-1,0)}.
Compute the values of x(t) and y(t) to obtain a parametrization of the real
points lying on the curve ² + y? = 1.
(b) Prove that x(t), y(t) are both rational numbers if and only if t is a rational
питber, and сoncude that
(x, y) E Q² | x² + y² = 1} = {(x(t), y(t)) | t e Q} U {(-1,0)}.
(c) Suppose now that we have a fundamental Pythagorean triple (a, b, c) so that
a, b, c e N, (a, b) = (a, c) = (b, c)
use Part (b) to prove that there exist relatively prime integers m > n > 1 of
opposite parity such that one of (a, b, c), (b, a, c) is equal to
1, and a? + b² = c?. Divide by c and
(2тп, т? — п?, т? + п?).
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