Question

DESPERATELY NEED PRECAL/TRIG HELP
I need help with parametric conics how to convert from parametric to rectangular form

- y = 1/8 (x-3)^2 + 5
x = 2 (y + 7)^2 -8

Answer 1

relating to cos and sin

Answer 2

HI!!

Answer 3

these look like they are already in rectangular form

Answer 4

Well then I guess convert it the other way? So it's like _ cos (t) _

Answer 5

and sin (t)

Answer 6

THIS IS JUST A EXAMPLE ON HOW TO DO IT
Graphic Representation of the Equations :

3y + 4x = 29 -3y + 2x = 1



Solve by Substitution :

// Solve equation [2] for the variable x


[2] 2x = 3y + 1

[2] x = 3y/2 + 1/2
// Plug this in for variable x in equation [1]

[1] 4•(3y/2+1/2) + 3y = 29
[1] 9y = 27
// Solve equation [1] for the variable y


[1] 9y = 27

[1] y = 3
// By now we know this much :

x = 3y/2+1/2
y = 3
// Use the y value to solve for x

x = (3/2)(3)+1/2 = 5
Solution :

{x,y} = {5,3}

Processing ends successfully

Answer 7

ok lets look it up

Answer 8

\[x = 2 (y + 7)^2 -8\] is a parabola

Answer 9

opens to the right

Answer 10

the way to write this conic section in parametric form is \[r=\frac{ed}{1-\cos(\theta)}\]

Answer 11

\(e\) is the eccentricity, which for a parabola is \(1\)

Answer 12

and \(d\) is the directrix

Answer 13

THIS IS JUST A EXAMPLE ON HOW TO DO IT
Graphic Representation of the Equations :

3y + 4x = 29 -3y + 2x = 1



Solve by Substitution :

// Solve equation [2] for the variable x


[2] 2x = 3y + 1

[2] x = 3y/2 + 1/2
// Plug this in for variable x in equation [1]

[1] 4•(3y/2+1/2) + 3y = 29
[1] 9y = 27
// Solve equation [1] for the variable y


[1] 9y = 27

[1] y = 3
// By now we know this much :

x = 3y/2+1/2
y = 3
// Use the y value to solve for x

x = (3/2)(3)+1/2 = 5
Solution :

{x,y} = {5,3}

Processing ends successfully