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

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)