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# p5 june 2004 help ! watch

1. Part B

can someone explain plz ?
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2. Doc1.doc (53.0 KB, 110 views)
3. Sub the equation of the normal back into xy = c²:

=> t²x² + (c/t)x - (ct³)x = c²

t³ + (c-ct4)x - tc² = 0

Now use the quadratic equation to solve for x:

x = {(ct4-c) ± √[(c-ct4)² + 4c²t4] } / 2t³

If you simplify the bit in the root, you get: √[c²(t8+2t4+1)]

= √[c²(t4+1)²]

= ct4 + c

So, x = [ct4-c + (ct4+c)]/2t³ or x = [ct4-c - (ct4+c)]/2t³

After simplification,

x = ct or x = -c/t³

x = ct is for the point P, therefore x = -c/t³ is for the point Q.

When x = -c/t³ , (-c/t³)y = c²

y = -ct³

Therefore, Q( -c/t³ , -ct³ )
4. thanks (bloody hell thats long ) do u know wat u need to get A grade on that paper ???? out of 75 ?????

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