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viva [34]
3 years ago
13

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

Mathematics
2 answers:
KiRa [710]3 years ago
6 0

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

sukhopar [10]3 years ago
5 0

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

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