Rewrite the parametric system as one equation. x = 2-4\tan(2t) y = \sec(2t) - 3

Question

Rewrite the parametric system as one equation.


`x = 2-4\tan(2t)`

`y = \sec(2t) - 3`

Answer 1

We can rewrite the equation for $x$ as

`x-2=-4\tan(2t)which \: gives \: us \\ (x-2)/(-4)=\tan(2t)`

We can also rewrite the equation for $y$ as

`y=\sec(2t) - 3 \: which \: gives \: us \: \\ y+3=\sec(2t).`

`Since \tan and \sec are \: related, we \: know \: that \tan^2 \\ +1=\sec^2. \: Therefore, \tan(2t)^2+1=\sec(2t)^2`

`Plugging \: this \: in, we \: get \tan \\ (2t)^2+1=(y+3)^2/(-4)^2.`

`Since \tan^2 \\ +1=\sec^2, we \: have (x-2)^2/(-4)^2=(y+3)^2/(-4)^2.`

`Simplifying \: further, we \: get \\ (x-2)^2=(y+3)^2.`

Answer 2

`(y+3)^2-((x-2)^2)/(16)=1`

Explanation:

You want the system of parametric equations (x, y) = (2-4tan(2t), sec(2t) -3) rewritten as one equation in x and y.

Trig identity

We can solve for tan(2t) and sec(2t) and use the trig identity ...

sec(2t)² = tan(2t)² +1

Trig expressions

y +3 = sec(2t) . . . . . . add 3

x -2 = -4tan(2t) . . . . . subtract 2

(x -2)/(-4) = tan(2t) . . . . divide by 4

Using these expressions in the above identity, we get ...

`(y+3)^2 = \left((x-2)/(-4)\right)^2 +1`

Subtracting the x-term gives the equation of a hyperbola that opens along the y-axis:

`\boxed{(y+3)^2-((x-2)^2)/(16)=1}`

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