parametric equation

Question

A curve has a parametric equation x = 2t − 4, y = 4t

2

. Find the

(a) cartesian equation of the curve;

(b) coordinates of the points where the line y = 4x + 28 meets the

curve.

Answer 1

x=2t-4

y = 4t

2t =y/2

x=y/2 -4

y=4x+28

y =4(y/2-4)+28 = 2y-16+28

y = -12

x= y/2-4 = -12/2-4 = -10

(x,y) = (-10,-12) is the point where the two lines meet

2x-y +8 = 0 is the cartesian equation for the two parametic equations

or in slope intercept form: y=2x+8

UNLESS you meant y=4t^2

then

x=2t-4

2t=x+4

t =x/2 +2

y=4(x/2+2)^2 = 4(x^2/4+2x+4) = x^2 +8x+16 = 4x + 28

x^2 + 4x -12=0

(x+6)(x-2) = 0

x=2, -6

y = 36, 4

(x,y) = (2,36) or (-6,4) as the intersection of the line and curve