How to find the cartesian equation from parametric equation?

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

2 Answers
Dec 15, 2017

x=y^2/16x=y216

Explanation:

We know that x=4t^2x=4t2 and y=8ty=8t. We're going to eliminate the parameter tt from the equations.

Since y=8ty=8t we know that t=y/8t=y8.

We can now substitute for tt in x=4t^2x=4t2:

x=4(y/8)^2\rightarrow x=(4y^2)/64\rightarrow x=y^2/16x=4(y8)2x=4y264x=y216

Although it is not a function, x=y^2/16x=y216 is a form of the Cartesian equation of the curve. It's frequently the case that you do not end up with yy as a function of xx when eliminating the parameter from a set of parametric equations.

Dec 15, 2017

"see explanation"see explanation

Explanation:

(a)(a)

y=8trArrt=1/8yy=8tt=18y

rArrx=4t^2=4xx(1/8y)^2=1/16y^2x=4t2=4×(18y)2=116y2

rArrx=1/16y^2larrcolor(blue)"cartesian equation"x=116y2cartesian equation

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

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

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

"the equation of the line passing through"the equation of the line passing through

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

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

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

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

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

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)xd=(x2x1)2+(y2y1)2

rArrd=sqrt((36-4)^2+(-24-8)^2)d=(364)2+(248)2

color(white)(rArrd)=sqrt2048=32sqrt2d=2048=322