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FP1

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
(edited 7 years ago)
Reply 1
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B_9710
18
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.
Reply 2
Avatar for Zacken
Zacken
22
Sketchity sketch sketch.
Reply 3
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B_9710
18
Also how did you get y2=16x?
Reply 4
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Zacken
22
Original post by B_9710
Also how did you get y2=16x?


Presumably: (4,8)(4, 8) lies on the parabola, so: 82=4a(4)⇒a=6416=4⇒y2=4ax=4(4)x=16x8^2 = 4a(4) \Rightarrow a = \frac{64}{16} = 4 \Rightarrow y^2 = 4ax = 4(4)x = 16x
Reply 5
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B_9710
18
Original post by Zacken
Presumably: (4,8)(4, 8) lies on the parabola, so: 82=4a(4)⇒a=6416=4⇒y2=4ax=4(4)x=16x8^2 = 4a(4) \Rightarrow a = \frac{64}{16} = 4 \Rightarrow y^2 = 4ax = 4(4)x = 16x


It says P(4,80)? Presumably a mistake.
Original post by B_9710
It says P(4,80)? Presumably a mistake.


Probably not a mistake, just sub y=80y=80 and x=4x=4 in to get a=400a=400 ?

EDIT: Ignore if you mean a mistake in Zacken's post
(edited 7 years ago)
Reply 7
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B_9710
18
Original post by IrrationalRoot
Probably not a mistake, just sub y=80y=80 and x=4x=4 in to get a=400a=400 ?


That's what I'm saying, the point (4,80) does not lie on the curve y2 =16x.
Reply 8
Avatar for Zacken
Zacken
22
Original post by IrrationalRoot
Probably not a mistake, just sub y=80y=80 and x=4x=4 in to get a=400a=400 ?

EDIT: Ignore if you mean a mistake in Zacken's post


Uhm, okay - so a=400a=400 - how does that make y2=16xy^2 = 16x?

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 ).
(edited 7 years ago)
Original post by Zacken
Uhm, okay - so a=400a=400 - how does that make y2=16xy^2 = 16x?


Oh sorry I saw those equations but I didn't really have a clue where they came from.
Reply 10
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Zacken
22
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. :tongue:
Original post by Zacken
See the edit to my post. :tongue:


K makes sense.
Reply 12
Avatar for Acrux
Acrux
OP
7
Original post by Zacken
Uhm, okay - so a=400a=400 - how does that make y2=16xy^2 = 16x?

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)
Reply 13
Avatar for Acrux
Acrux
OP
7
So the sketch suggest it can only be in the 4th quadrant is that the reason why?
thanks in advance
Original post by Zacken
Sketchity sketch sketch.
Reply 14
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B_9710
18
Well, firstly we know that point Q, x≠4 x\neq4 as x=4 at point P which is already given. So x=36.
Reply 15
Avatar for Acrux
Acrux
OP
7
Original post by B_9710
Well, firstly we know that point Q, x≠4 x\neq4 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
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)
(edited 7 years ago)
Reply 17
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Acrux
OP
7
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..

Read the post again.
And anyway this thread has been answered

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