Posted by **TRACY-888** 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)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

y=(1/2)(p+q)x-apq

That is not the equation of a parabola.

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