Proving With A Tangent, A Parabola and Degree

[sadness]

This question is tricky so I hope someone can help me understand this question, correct me and in the end, prove it.

The point P(a,b) is on the parabola x^2 =4y . The tangent at P meets the line y=-1 at the
point A. For the point F(O,1) ,prove that (angle)<AFP=90° for all positions P, except (0,0).

What I got so far:

y = 1/4x^2 and y' = 1/2x
Let T be (c, -1)

To prove that it has a 90 degrees, I thought that we have to prove that a^2 + b^2 = c^2 which is essentially the distance of FP^2 + d of FT^2 = d of PT^2

With that I started to use (a,b) to find value of c which I need the line of tangent

y' = 1/2 (a) <--- slope

y - b = 1/2a (x - a)

y= 1/2ax - 1/2a + b < ----- line of tangent at (a,b)

Substituting -1 to the line of tangent to get c value

(-1) = 1/2 ax - 1/2a + b
c= x = 2(1/2a -b +1) / a

So I now have F(1,0) , P(a,b), T(-1, (a-2b+2)/a), Three points that I can use to find the distance to get line a, b, c to prove.

Line equation: sqrt[(x2-x1)^2 + (y2 - y1)^2]



Distance of P to F= sqrt [ (a-1)^2 + (b)^2 ] (order wouldn't matter right?

Distance of F to T = sqrt [ (1-(-1))^2 + (a-2b+2/a)^2]
= sqrt [ 4 + (a-2b+2/a)^2 ]

Distance of P to T= sqrt [ (a-(-1)^2 + ((a-2b+2/a) - b)^2]

Technically speaking, if each distance is getting squared (a^2 + b^2 = c^2), that would cancel the sqrt for each distance.

Now.... I'm confused from here

[(1-a)^2 + (-b)^2] + [4 + (a-2b+2/a)^2] = (a+1)^2 + ((a-2b+2/a) - b) ^2

1- 2a + a^2 + 4 + (a-2b+2/a)^2 = a^2 + 2a + 2 + ((a-2b+2/a)-b)^2
a^2 - 2a +5 + (a-2b+2)^2/a^2 = a^2 + 2a + 2 + ((a-2b+2/a)-b)^2

And stuck now. Please point out where I'm wrong!

If there is another way to do this question, please do tell!

Thank you

 

[DrPhil]

This question is tricky so I hope someone can help me understand this question, correct me and in the end, prove it.

The point P(a,b) is on the parabola x^2 =4y . The tangent at P meets the line y=-1 at the
point A. For the point F(O,1) ,prove that (angle)<AFP=90° for all positions P, except (0,0).

What I got so far:

y = (1/4)x^2 and y' = (1/2)x ... Use more parentheses when typing inline
Let T be (c, -1) ......Is point T the same as A in the problem?

To prove that it has a 90 degrees, I thought that we have to prove that a^2 + b^2 = c^2 which is essentially the distance of FP^2 + d of FT^2 = d of PT^2

With that I started to use (a,b) to find value of c which I need the line of tangent

y' = (1/2) (a) <--- slope

y - b = (1/2)a (x - a)

y= (1/2)ax - (1/2)a^2 + b < ----- line of tangent at (a,b)
Error here .. a^2
BUT b = (a^2)/4, so you can simplify further
.......y = (a/2)x + (a^2)/4

Substituting -1 to the line of tangent to get c value

(-1) = 1/2 ax - 1/2a + b
c= x = 2(1/2a -b +1) / a

So I now have F(1,0) , P(a,b), T(-1, (a-2b+2)/a), Three points that I can use to find the distance to get line a, b, c to prove. OOPS - point F is (0,1), NOT (1,0) !!!

Line equation: sqrt[(x2-x1)^2 + (y2 - y1)^2]


Distance of P to F= sqrt [ (a-1)^2 + (b)^2 ] (order wouldn't matter right?
..................|PF| = sqrt[ a^2 + (b - 1)^2]

Distance of F to T = sqrt [ (1-(-1))^2 + (a-2b+2/a)^2]
= sqrt [ 4 + (a-2b+2/a)^2 ]

Distance of P to T= sqrt [ (a-(-1)^2 + ((a-2b+2/a) - b)^2]

Technically speaking, if each distance is getting squared (a^2 + b^2 = c^2), that would cancel the sqrt for each distance. Yes - easier to work with squares of distances.

Now.... I'm confused from here
. . .

If there is another way to do this question, please do tell!

Thank you

Thank you for showing your work!

When typing formulas inline, you have to use enough parentheses that the order of operations is clear and unambiguous.

The line y=-1 is called the "directrix" of the parabola, and point F=(0,1) is the "focus." If you have studied those terms, you might be able to use them in the proof - but what you are doing should be good - IF you fix the several errors!

If you still need help, show us how far you have gotten after fixing.

 

[sadness]

Thank you for showing your work!

When typing formulas inline, you have to use enough parentheses that the order of operations is clear and unambiguous.

The line y=-1 is called the "directrix" of the parabola, and point F=(0,1) is the "focus." If you have studied those terms, you might be able to use them in the proof - but what you are doing should be good - IF you fix the several errors!

If you still need help, show us how far you have gotten after fixing.

Thank you for correcting my errors!
Unfortunately I still have problems solving it in the end so here what I have so far after correcting it
I want to apologize in advance if my brackets are not in the right place

line of tangent
y= (a/2)x + (a^2 / 4)

sub in -1 to line of tangent to get c
(-1) = (a/2)x + (a^2 / 4)
-[(a/2)x] = (a^2 / 4) -1
x = c = -(a^2 / 2) - 2a

sub in distance formula F (0,1), P (a, a^2/4) and A (-(a^2 /2) - 2a , -1 )

FP = sqrt [ a^2 + (b-1)^2 ]
FA = sqrt [ (0-((-a^2 /2) -2))^2 + (1-(-1)^2]
= sqrt [ (a^2/2 + 2a)^2 + 4)
PA = sqrt[ a -((-a^2 /2) -2))^2 + (b-(-1))^2]
= sqrt [(a^2/2 +3a) + (b+1)^2 ]

a^2 + b^2 = c^2
(FP)^2 + (FA)^2 = (PA)^2
took out the sqrt and sub in b = a^2 / 4

a^2 + ((a^2/4) - 1)^2 + (a^2/2 + 2a)^2 + 4 = (a^2/2 + 3a) + ((a^2/4) +1))
Please tell me this end equation does equal to each other in the very end.
When I expand and simplify, it doesn't

Thank you!

 

[DrPhil]

Thank you for correcting my errors!
Unfortunately I still have problems solving it in the end so here what I have so far after correcting it
I want to apologize in advance if my brackets are not in the right place

line of tangent
y= (a/2)x + (a^2 / 4)

sub in -1 to line of tangent to get c
(-1) = (a/2)x + (a^2 / 4)
-[(a/2)x] = (a^2 / 4) -1 ..... OH-OH .. (-1) on the left becomes (+1) on the right
x = c = -(a^2 / 2) - 2a....... Several errors in this step - should be
......c = - (a/2) - (2/a)
sub in distance formula F (0,1), P (a, a^2/4) and A (-(a^1 /2) - 2/a , -1 )

FP = sqrt [ a^2 + (b-1)^2 ]
FA = sqrt [ (0-((-a^2 /2) -2))^2 + (1-(-1)^2]
= sqrt [ (a^2/2 + 2a)^2 + 4)
PA = sqrt[ a -((-a^2 /2) -2))^2 + (b-(-1))^2]
= sqrt [(a^2/2 +3a) + (b+1)^2 ]

a^2 + b^2 = c^2
(FP)^2 + (FA)^2 = (PA)^2
took out the sqrt and sub in b = a^2 / 4

a^2 + ((a^2/4) - 1)^2 + (a^2/2 + 2a)^2 + 4 = (a^2/2 + 3a) + ((a^2/4) +1))
Please tell me this end equation does equal to each other in the very end.
When I expand and simplify, it doesn't

Thank you!

Hope it works out better with point A corrected!