locus (1 Viewer)

[super.muppy]

find a point on the parabola x^2=4ay that the normal passes through its focus

this question raped my mind

 

[life92]

x^2=4ay
Focus (0,a)

y = x^2 / 4a
y' = x / 2a
At some point (2ap,ap^2)
y' = 2ap / 2a
= p

Therefore, gradient of normal = -1/p
Eqn of normal: y -ap^2=-1/p(x-2ap)
py-ap^3=-x+2ap
x+py=ap^3+2ap

Through (0,a) (focus)
ap = ap^3 + 2ap
0 = ap^3 + ap
0 = ap (p^2+1)
Therefore, p = 0 only

That means that the only normal that passes through the focus, is the normal through the origin (0,0), that is, the line x=0

Hope that helps !

 

[super.muppy]

x^2=4ay
Focus (0,a)

y = x^2 / 4a
y' = x / 2a
At some point (2ap,ap^2)
y' = 2ap / 2a
= p

Therefore, gradient of normal = -1/p
Eqn of normal: y -ap^2=-1/p(x-2ap)
py-ap^3=-x+2ap
x+py=ap^3+2ap

Through (0,a) (focus)
ap = ap^3 + 2ap
0 = ap^3 + ap
0 = ap (p^2+1)
Therefore, p = 0 only

That means that the only normal that passes through the focus, is the normal through the origin (0,0), that is, the line x=0

Hope that helps !

true, but is there a way without parameters namely (2ap,ap^2)? (cuz i saw this in a 2U) book yea i noe wrong forum.

 

[super.muppy]

and also p=0 ? what happens if say p=2?

 

[life92]

You just think your SOOO tuff with that 99 ext 1 mark, how about i 99 ext 1 mark your face.
Just kidding, what are you studying?

lol wth dude.
I'm just trying to help out...........

But in answer to super.muppy's question about not using parameters, I tried it again and the answer doesn't seem to come out which is pretty weird.

I think you could possibly think of this problem as a non-algebraic problem and try to think about it logically. So it then becomes obvious that the tangent at the origin is horizontal and the normal would then be the y axis and hence pass through the focus.

Also, if p=2, and we know that the equation of the normal is x+py=ap^3+2ap, then x+2y = 8a+4a
x+2y=12a
Now the focus is still (0,a), and if you sub that into the equation, LHS = 2a, which does NOT equal the RHS and so does not pass through the focus.

 

[Lukybear]

Hm... life using algebra

did u get y=a?

 

[life92]

Hm... life using algebra

did u get y=a?

Oh my bad. I made a very silly mistake lol.

x^2 = 4ay
y = x^2 / 4a
y' = x/2a
At some point (x1,y1)
y' = x1/2a

Therefore, the gradient of the normal = -2a / x1

y-y1=-2a / x1 (x-x1)
Rearranging: 2ax + x1 y = x1 y1 + 2a x1
Subbing (0,a)
0 + x1 a = x1 y1 + 2a x1
0 = x1 y1 + x1 a
0 = x1 (y1 + a)

Now, x1 = 0 or y1 = -a, but y=-a does not lie on the parabola x^2=4ay, and hence x1 = 0, which then gives you the final answer as the origin.

 

[johNNyboy772]

lol @ raped your mind

hahahaha thats pretty funny