Question

Eliminate the parameter and obtain the standard form of the rectangular equation. Ellipse: x = h + a cos(θ), y = k + b sin(θ) Use your result to find a set of parametric equations for the line or conic. (When 0 ≤ θ ≤ 2π. Enter your answers as a comma-separated list of equations.) Ellipse: vertices: (−4, 0), (6, 0); foci: (−2, 0), (4, 0)

Answer 1

Answer:

a)  $(\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1$

b)  $x=1+5 \,cos(\theta)\,,\,y=4\,sin(\theta)$

c)   $\frac{(x-1)^2}{25}+ \frac{y^2}{16}=1$

Step-by-step explanation:

Start by isolating the trigonometric expression in both equations, ad then use the Pythagorean identity:

$cos^2(\theta)+sin^2(\theta)=1$

to obtain the standard equation of a conic.

$x=h+a\,cos(\theta)\\x-h=a\,cos(\theta)\\cos(\theta)=\frac{x-h}{a}$                   $y=k+b\,sin(\theta)\\y-k=b\,sin(\theta)\\sin(\theta)=\frac{y-k}{b}$

then:

$cos^2(\theta)+sin^2(\theta)=1\\(\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1$

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:

$x=h+a \,cos(\theta)\,,\,y=k+b\,sin(\theta)$ 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 $a=5$. The center of the ellipse is therefore at (1,0).

We recall that the vertices of a translated horizontal ellipse are located at $(a+h,k)$ and $(-a+h,k)$, then k=0 to satisfy the information given (-4,0) & (6,0), and since $a=5$, we deduce that $a+h = 6$  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: $c=\sqrt{a^2-b^2}$

$c=\sqrt{a^2-b^2}\\3=\sqrt{5^2-b^2}\\9=25-b^2\\b^2=25-9\\b^2=16\\b=4$

Then we can write the equation of this ellipse in standard form as:

$(\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1\\(\frac{x-1}{5})^2+ (\frac{y-0}{4})^2=1\\\\(\frac{x-1}{5})^2+ (\frac{y}{4})^2=1\\\frac{(x-1)^2}{25}+ \frac{y^2}{16}=1$