Answer:
a) [tex](\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1[/tex]
b) [tex]x=1+5 \,cos(\theta)\,,\,y=4\,sin(\theta)[/tex]
c) [tex]\frac{(x-1)^2}{25}+ \frac{y^2}{16}=1[/tex]
Step-by-step explanation:
Start by isolating the trigonometric expression in both equations, ad then use the Pythagorean identity:
[tex]cos^2(\theta)+sin^2(\theta)=1[/tex]
to obtain the standard equation of a conic.
[tex]x=h+a\,cos(\theta)\\x-h=a\,cos(\theta)\\cos(\theta)=\frac{x-h}{a}[/tex] [tex]y=k+b\,sin(\theta)\\y-k=b\,sin(\theta)\\sin(\theta)=\frac{y-k}{b}[/tex]
then:
[tex]cos^2(\theta)+sin^2(\theta)=1\\(\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1[/tex]
this is the equation of an ellipse centered at (h,k), and with horizontal axis length = 2a , and vertical axis length = 2b.
The parametric equations are those we started with:
[tex]x=h+a \,cos(\theta)\,,\,y=k+b\,sin(\theta)[/tex] but we need to find the appropriate parameters for the requested ellipse, as shown below.
For an ellipse of vertices (-4,0) a,d (6,0) and foci at (-2,0) and (4.0), we are dealing with an ellipse with major horizontal axis on the line y=0, and major diameter length of 10 units, so the parameter [tex]a=5[/tex]. The center of the ellipse is therefore at (1,0).
We recall that the vertices of a translated horizontal ellipse are located at [tex](a+h,k)[/tex] and [tex](-a+h,k)[/tex], then k=0 to satisfy the information given (-4,0) & (6,0), and since [tex]a=5[/tex], we deduce that [tex]a+h = 6[/tex] and therefore h=1.
To find "b" (the only parameter missing for the standard equation of the conic), we need the information on the foci (-2,0) and (4,0) which must equal (h-c,k) and (h+c,k) with k=0 and h=1 which then gives that c=3
Now using the formula for the parameter "c" of the foci: [tex]c=\sqrt{a^2-b^2}[/tex]
[tex]c=\sqrt{a^2-b^2}\\3=\sqrt{5^2-b^2}\\9=25-b^2\\b^2=25-9\\b^2=16\\b=4[/tex]
Then we can write the equation of this ellipse in standard form as:
[tex](\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1\\(\frac{x-1}{5})^2+ (\frac{y-0}{4})^2=1\\\\(\frac{x-1}{5})^2+ (\frac{y}{4})^2=1\\\frac{(x-1)^2}{25}+ \frac{y^2}{16}=1[/tex]
