Eliminating the parameter from a parametric equation

There are three ways to eliminate the parameter from a parametric equation

Given a parametric curve where our function is defined by two equations, one for ???x??? and one for ???y???, and both of them in terms of a parameter ???t???,

???x=f(t)???

???y=g(t)???

we can eliminate the parameter value in a few different ways.

We can

1. solve each equation for the parameter ???t???, then set the equations equal to one another, or

2. solve one equation for the parameter ???t???, then plug that value into the second equation, or

3. solve each equation for part of an identity, then plug both values into the identity.

## Eliminating the parameter using the second method

Example

Eliminate the parameter.

???x=2t^2+6???

???y=5t???

We’ll solve ???y=5t??? for ???t???, since this will be easier than solving ???x=2t^2+6??? for ???t???.

???y=5t???

???t=\frac{y}{5}???

Plugging this into the equation for ???x???, we get

???x=2\left(\frac{y}{5}\right)^2+6???

???x=\frac{2y^2}{25}+6???

Removing the fraction, we get

???25x=2y^2+150???

???25x-2y^2=150???

Given a parametric curve where our function is defined by two equations, one for x and one for y, and both of them in terms of a parameter t, we can eliminate the parameter in a few different ways.

Example

Eliminate the parameter.

???x=e^t???

???y=e^{4t}???

We know that ???y=e^{ab}??? is the same as ???y=(e^a)^b???. If we use this property, we can take ???y=e^{4t}??? and rewrite it as ???y=(e^t)^4???. Since ???x=e^t???, we can substitute ???x??? into ???y=(e^t)^4??? for ???e^t???.

???y=x^4???

Remember, because we have ???e??? in the original parametric equations, and ???e??? requires that ???t>0???, we have to transfer this condition to our final answer, and say

???y=x^4???, where ???x>0???

Let’s try another example using the third method.

Example

Eliminate the parameter.

???x=2\cos{\theta}???

???y=3\sin{\theta}???

???0\le\theta\le2\pi???

Rearranging ???x=2\cos{\theta}??? and ???y=3\sin{\theta}??? to isolate the trigonometric functions, we get

???x=2\cos{\theta}???

???\cos{\theta}=\frac{x}{2}???

and

???y=3\sin{\theta}???

???\sin{\theta}=\frac{y}{3}???

Since we know that ???\sin^2{\theta}+\cos^2{\theta}=1???, we can substitute the values we just found for ???\cos{\theta}??? and ???\sin{\theta}???.

???\left(\frac{y}{3}\right)^2+\left(\frac{x}{2}\right)^2=1???

???\frac{y^2}{9}+\frac{x^2}{4}=1???

???y^2+\frac{9x^2}{4}=9???

???4y^2+9x^2=36???

???9x^2+4y^2=36???

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