Question

Please help!! x = 2 - 3 cos t, y = 1 + 4 sin t in rectangular form? thank you! :)

Answer 1

Answer:

Option 1

Step-by-step explanation:

x = 2 - 3cost

cost = (x-2)/-3

(cost)² = (x-2)²/9

y = 1 + 4sint

sint = (y-1)/4

(sint)² = (y-1)²/16

cos² + sin² = 1

(x-2)²/9 + (y-1)²/16 = 1

Answer 2

Answer:

$\frac{(x-2)^2}{9}+\frac{(y-1)^2}{16}=1$

Step-by-step explanation:

$x=2-3\cos(t)$

$y=1+4\sin(t)$

Let's solve for $\cos(t)$ in the first equation and then solve for $\sin(t)$ in the second equation.

I will then use the following identity to get right of the parameter, $t$:

$\cos^2(t)+\sin^2(t)=1$ (Pythagorean Identity).

Let's begin with $x=2-3\cos(t)$.

Subtract 2 on both sides:

$x-2=-3\cos(t)$

Divide both sides by -3:

$\frac{x-2}{-3}=\cos(t)$

Now time for the second equation, $y=1+4\sin(t)$.

Subtract 1 on both sides:

$y-1=4\sin(t)$

Divide both sides by 4:

$\frac{y-1}{4}=\sin(t)$

Now let's plug it into our Pythagorean Identity:

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

$\frac{x-2}{-3})^2+(\frac{y-1}{4})^2=1$

$\frac{(x-2)^2}{(-3)^2}+\frac{(y-1)^2}{4^2}=1$

$\frac{(x-2)^2}{9}+\frac{(y-1)^2}{16}=1$