Graph the line with slope 3/4 passing through the point (-1, 2).

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

sukhopar [10]

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:

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

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

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

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

5 0

3 years ago

Read 2 more answers

For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?

Bond [772]

Given that $r=\theta$ , then $r'=1$

The slope of a tangent line in the polar coordinate is given by:

$m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$

Thus, we have:

$m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}$

Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

$\sin\theta+\theta\cos\theta=0 \\ \\ \theta\cos\theta=-\sin\theta \\ \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

θ = 0

θ = 2.02875783811043

θ = 4.91318043943488

Part B:

For vertical tangent lines, $\frac{1}{m} =0$

Thus, we have:

$\cos\theta-\theta\sin\theta=0 \\ \\ \Rightarrow\theta\sin\theta=\cos\theta \\ \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

θ = 4.91718592528713

3 0

3 years ago

Find the vertical asymptote(s) of f of x equals quantity 3 x squared plus 3x plus 6 end quantity over quantity x squared minus 2

Salsk061 [2.6K]

Answer:

a i think give me brainliest. good day.

Step-by-step explanation:

7 0

3 years ago

No anyone know these answers ?

Pepsi [2]

Answer: Positive Quadratic