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# Realtive to C4 ..please help ;) watch

1. Hey TSR-ers! Please, help me out with this question from edexcel core mathematics 4 textbook (i do not have the solution bank ..if anyone has the link to it please link me with B-) thanks) !

From chapter 2, exercise 2B

Q6. A curve is given parametrically by the equations x=3qt^2, y=4(t^3+1) , where q is a constant. The curve meets the x-axis at X and the y-axis at Y. Given that OX=2OY, where O is the origin. Find the value of q.

[ANSWER: 8/3]

Help appreciated & thanks xD but reply needed asap
2. It meets the x-axis at X, which is precisely when y=0, and so setting y=0 and x=X you can find an expression for X in terms of q (by solving the equation y=0 for t and substituting that value into the x equation). You do a similar thing for Y, by setting x=0 and so on, to find an expression for Y in terms of q, and then use the fact that |X|=2|Y| to find the value of q.
3. first thing you need to do is find the points where the x,y coordinates of X and Y. X is on the x axis, so its y-coordinate is 0. so solve y(t)=0 for t, then subs your value of t into x(t). that'll give you your x co-ordinate of X. Same idea for Y, just solve x(t)=0. Since they're both on the axes, the length of OX is just the x-coordinate of x, and length of OY is the y co-ordinate at Y. You'll have both of these lengths (or just 1) in terms of q, so just solve OX=2OY.
4. (Original post by bluefire)
Hey TSR-ers! Please, help me out with this question from edexcel core mathematics 4 textbook (i do not have the solution bank ..if anyone has the link to it please link me with B-) thanks) !

From chapter 2, exercise 2B

Q6. A curve is given parametrically by the equations x=3qt^2, y=4(t^3+1) , where q is a constant. The curve meets the x-axis at X and the y-axis at Y. Given that OX=2OY, where O is the origin. Find the value of q.

[ANSWER: 8/3]

Help appreciated & thanks xD but reply needed asap

LET Y = 0 /* find where curve meets x- axis */

4 (t^3 +1 ) = 0

t = -1^1/3 /* substitute into x equation */

x = 3q (-1^1/3)^2 = 3q OX = 3q

LET X = 0 /* find where curve meets y- axis */

3qt^2 = 0

t= 0 /* substitute into y equation */

y = 4( 0 + 1) = 4 OY = 4

OX = 2OY

3q = 2 * 4
3q = 8

q = 8/3
5. done =D thanks my friend ! i guess i had a mind block lol !
6. (Original post by kaosu_souzousha)
Spoiler:
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LET Y = 0 /* find where curve meets x- axis */

4 (t^3 +1 ) = 0

t = -1^1/3 /* substitute into x equation */

x = 3q (-1^1/3)^2 = 3q OX = 3q

LET X = 0 /* find where curve meets y- axis */

3qt^2 = 0

t= 0 /* substitute into y equation */

y = 4( 0 + 1) = 4 OY = 4

OX = 2OY

3q = 2 * 4
3q = 8

q = 8/3
It's generally a lot more helpful to let the person work the answer out for themselves (with nudges in the right direction) rather than giving them the full answers. If you do insist on putting full solutions in, at the very least put them within spoiler tags. Refer to this for more on the subject of full answers:
http://www.thestudentroom.co.uk/showthread.php?t=403989
7. (Original post by immense010)
It's generally a lot more helpful to let the person work the answer out for themselves (with nudges in the right direction) rather than giving them the full answers. If you do insist on putting full solutions in, at the very least put them within spoiler tags. Refer to this for more on the subject of full answers:
http://www.thestudentroom.co.uk/showthread.php?t=403989
Thank you for pointing me in the right direction.
If I seem to give a damn feel free to tell me. I would hate to be giving the wrong impression.

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