For parametric equations x= a cos t and y= b sin t, describe how the values of a and b determine which conic section will be traced
Question
Answer:
In mathematics, a conic section is
a curve obtained as the intersection of the surface of
a cone with a plane. The four types of conic section are
the hyperbola, the parabola, the circumference and the ellipse.For the problem we have this parametric equation:
(1) [tex]\left\{{{x=acost}\atop{y=bsint}}\right[/tex]
From geometry, we know that we can express a circumference in terms of parameters like this:
(2) [tex]\left \{ {{x=rcost} \atop {y=rsint}}\right[/tex]
being r the radius of the circumference.
On the other hands, we know that a ellipse can be expressed in terms of parameters like this:
(3) [tex]\left \{ {{x=acost} \atop {y=bsint}}\right[/tex]
Therefore, we will have three answers that are the cases for the values a and b, namely.
Case 1: Circumference
To the case of a circumference, the more simple ordinary equation is given by:
(4) [tex]x^{2} + y^{2} = r^{2}[/tex]
Substituting (1) into (4):
[tex]a^{2}cos^{2}t+b^{2}cos^{2}t=r^{2}[/tex]
But because of the equation (2), necessarily:
[tex]a = b = r[/tex]
Case 2: Ellipse (focal axis matches the x-axis)
In this case, the simple ordinary equation is given by:
(5) [tex]\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1[/tex]
being a and b semi-major axis and semi-minor axis respectively.
Given that a an b are variables of the parametrization, and a and b are variables of the ellipse as well, to avoid confusion we will modify the equation (5) like this:
(6) [tex]\frac{x^{2}}{a'^{2}}+\frac{y^{2}}{b'^{2}}=1[/tex]
So, substituting (2) into (6):
[tex]\frac{a^{2}cos^{2}t}{a'^{2}}+\frac{b^{2}cos^{2}t}{b'^{2}}=1[/tex]
Necessarily:
[tex]a=a'[/tex] and [tex]b=b'[/tex]
and given that the focal axis matches the x-axis, then:
[tex]a>b[/tex]
Case 3: Ellipse (focal axis matches the y-axis)
In this case, applying the same previous reasoning, the simple ordinary equation is given by:
(7) [tex]\frac{x^{2} }{b'^{2}}+\frac{y^{2}}{a'^{2}}=1[/tex]
being a' and b' semi-major axis and semi-minor axis respectively.
So, substituting (2) into (7):
[tex]\frac{a^{2}cos^{2}t}{b'^{2}}+\frac{b^{2}cos^{2}t}{a'^{2}}=1[/tex]
Necessarily:
[tex]a = b'[/tex] and [tex]b = a'[/tex]
and given that the focal axis matches the y-axis, then:
[tex]a<b[/tex]
Finally, the conclusions are:
1. If [tex]a = b[/tex] then a circumference will be traced. (See Figure 1)2. If [tex]a>b[/tex] then a ellipse will be traced with focal axis matching the x-axis. (See Figure 2)3. If [tex]a<b[/tex] then a ellipse will be traced with focal axis matching the y-axis. (See Figure 3)
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general
11 months ago
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