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)

a)  

b)  

c)  

Step-by-step explanation:

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

to obtain the standard equation of a conic.

then:

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:

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 . The center of the ellipse is therefore at (1,0).

We recall that the vertices of a translated horizontal ellipse are located at and , then k=0 to satisfy the information given (-4,0) & (6,0), and since , we deduce that  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:

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

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)