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# FP1

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1. P(4,80 lies on parabola y^2=4ax
the normal to c at p cuts the parabola again at point q fin coordinates of q

two equations
y=-x+12
y^2=16x

if you put y into y^2 you get x=36 and x=4 however at this point y does not equal =-24 to the equation y^2=16x
you get +- 24 the answer is -24 so in this question which value do i take
2. (Original post by Acrux)
P(4,80 lies on parabola y^2=4ax
the normal to c at p cuts the parabola again at point q fin coordinates of q

two equations
y=-x+12
y^2=16x

if you put y into y^2 you get x=36 and x=4 however at this point y does not equal =-24 to the equation y^2=16x
you get +- 24 the answer is -24 so in this question which value do i take
Think about the shape of the curve and where the normal would intersect at.
3. Sketchity sketch sketch.
4. Also how did you get y2=16x?
5. (Original post by B_9710)
Also how did you get y2=16x?
Presumably: lies on the parabola, so:
6. (Original post by Zacken)
Presumably: lies on the parabola, so:
It says P(4,80)? Presumably a mistake.
7. (Original post by B_9710)
It says P(4,80)? Presumably a mistake.
Probably not a mistake, just sub and in to get ?

EDIT: Ignore if you mean a mistake in Zacken's post
8. (Original post by IrrationalRoot)
Probably not a mistake, just sub and in to get ?
That's what I'm saying, the point (4,80) does not lie on the curve y2 =16x.
9. (Original post by IrrationalRoot)
Probably not a mistake, just sub and in to get ?

EDIT: Ignore if you mean a mistake in Zacken's post
Uhm, okay - so - how does that make ?

What I'm saying is that he meant to type P(4, 8) but instead he typod the last bracket and instead of pressing shift to enter the close bracket he typed P(4,80 where the 0 should be a ).
10. (Original post by Zacken)
Uhm, okay - so - how does that make ?
Oh sorry I saw those equations but I didn't really have a clue where they came from.
11. (Original post by IrrationalRoot)
Oh sorry I saw those equations but I didn't really have a clue where they came from.
See the edit to my post.
12. (Original post by Zacken)
See the edit to my post.
K makes sense.
13. (Original post by Zacken)
Uhm, okay - so - how does that make ?

What I'm saying is that he meant to type P(4, 8) but instead he typod the last bracket and instead of pressing shift to enter the close bracket he typed P(4,80 where the 0 should be a ).
(Original post by B_9710)
It says P(4,80)? Presumably a mistake.
yup a mistake P(4,8)
14. So the sketch suggest it can only be in the 4th quadrant is that the reason why?
(Original post by Zacken)
Sketchity sketch sketch.
15. Well, firstly we know that point Q, as x=4 at point P which is already given. So x=36.
16. (Original post by B_9710)
Well, firstly we know that point Q, as x=4 at point P which is already given. So x=36.
yes I know
however in y^2=16x y=+-24 when x=36
initial problem was knowing if it is negative or positive.. the answer was (36,-24)

But the sketch explains why
17. (Original post by Acrux)
P(4,80 lies on parabola y^2=4ax
the normal to c at p cuts the parabola again at point q fin coordinates of q

two equations
y=-x+12
y^2=16x

if you put y into y^2 you get x=36 and x=4 however at this point y does not equal =-24 to the equation y^2=16x
you get +- 24 the answer is -24 so in this question which value do i take
Why not get the equation of the normal then sub in y^2/16 in it. You will be able to get a quadratic formula to get the coordinates . When I did this I got Q(36,-24)
18. (Original post by Funnycatvideos)
Why not get the equation of the normal then sub in y^2/16 in it. You will be able to get a quadratic formula to get the coordinates . When I did this I got Q(36,-24)
What?
Thats what i did look at the equations i got..

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