Posted by TRACY888 on Monday, April 6, 2009 at 5:23am.
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)xapq 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
y=(1/2)(p+q)xapq
That is not the equation of a parabola.
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