Question

How to find the cartesian equation from parametric equation?

Can someone please explain to me how to do question 2? Thanks!

Answer 1

`x=y^2/16`

Explanation:

We know that `x=4t^2` and `y=8t`. We're going to eliminate the parameter `t` from the equations.

Since `y=8t` we know that `t=y/8`.

We can now substitute for `t` in `x=4t^2`:

`x=4(y/8)^2\rightarrow x=(4y^2)/64\rightarrow x=y^2/16`

Although it is not a function, `x=y^2/16` is a form of the Cartesian equation of the curve. It's frequently the case that you do not end up with `y` as a function of `x` when eliminating the parameter from a set of parametric equations.

Answer 2

`"see explanation"`

Explanation:

`(a)`

`y=8trArrt=1/8y`

`rArrx=4t^2=4xx(1/8y)^2=1/16y^2`

`rArrx=1/16y^2larrcolor(blue)"cartesian equation"`

`(b)color(white)(x)"substitute values of t into x and y"`

`t=1tox=4,y=8rArr(4,8)`

`t=-1tox=4,y=-8rArr(4,-8)`

`"the equation of the line passing through"`

`(color(red)(4),8)" and "(color(red)(4),-8)" is "x=4`

`(c)color(white)(x)" substitute values of t into x and y"`

`t=1tox=4,y=8rArr(4,8)`

`t=-3tox=36,y=-24rArr(36,-24)`

`"calculate the length using the "color(blue)"distance formula"`

`•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`rArrd=sqrt((36-4)^2+(-24-8)^2)`

`color(white)(rArrd)=sqrt2048=32sqrt2`