Find the equation of the chord of the parabola joining the points with parameters
(a)1 and -3 on x=2t, y=t^2
(b)1/2 and 2 on x=4t, y=2t^2
(c)-1 and -2 on x=t, y=1/2t^2
(d)-2 and 4 on x=1/2t^2, y=1/4t^2
Then use the formula y=1/2(p+q)x-apq to obtain the chords in parts (a)-(d)
PLEASE&THANKY0U VERY MUCH :)
3unit maths - drwls, Monday, April 6, 2009 at 6:36am
(d) is not a parabola since both x and y are proportional to the parameter t^2. Did you copy the problem correctly?
These are four separate problems that can be all done the same way. Consider (a):
t = x/2, so y = (1/4) x^2 is the equation of the parabola.
When t = 1, x = 2 and y = 1
When t = -3, x = -6 and y = 9
The chord that they want connects these two points.
The slope is 8/(-8) = -1
y = -x + b
1 = -2 + b
b = 3
y = -x + 3 is the chord equation, with
-6 < x < 2
You need to define a, p and q in the equation
That is not the equation of a parabola.