A ( 1 , 0 ) is a point on the parabola y = 2 x ( x − 1 ) . From point A , point P is moving along the curve towards the origin O ( 0 , 0 ) . What is the limit P → O lim sec ∠ A P O ?
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We can also use vectors to find the angle between vectors PA and PO.
The slope at the origin is − 2 (I used calculus to get that so I guess I sorta cheated). Let angle A O P = θ . Then
tan θ = − 2 . In this case θ is negative but we will take the positive value of theta for our purposes.
Note that as point P approaches point O ∠ P A O approaches 0 . This means that the limiting value of ∠ A P O = 1 8 0 − θ , call it α . We may realize that
tan α = 2 − 1 .
Now we may construct a right triangle with sides 2 and 1 and hypotenuse equal to 5 (by the Pythagorean Theorem ), and from here it is simple to deduce that sec α = 5 − 1 .
Recall: The slope m ( P O ) of the secant line P O is the average rate of change between P and O lim P → O m ( P O ) = Slope of the tangent line at point O
What the problem is essentially asking is the secant of the angle between the tangent line at point O and the line O A .
To solve the problem we need to find the angle θ using the dot product of T and O A . To do this we must first find vector T .
m ( x ) = d x d ( 2 x ( x − 1 ) ) = 4 x − 2 and m ( 0 ) = − 2 T = ⟨ − 1 , 2 ⟩ and O A = ⟨ 1 , 0 ⟩ Note: To find vector T just choose two points on the line y = -2x, e.g (-1,2) and (0,0), and subtract their components.
Now that we have the two vectors we can solve the problem.
T ∙ O A = − 1 ∣ T ∣ = 5 ∣ O A ∣ = 1 T ∙ O A = ∣ T ∣ ∣ O A ∣ c o s ( θ ) → c o s ( θ ) = ∣ T ∣ ∣ O A ∣ T ∙ O A = 5 − 1
Finally: s e c ( θ ) = c o s ( θ ) 1 = − 5
My solution involves finding ∠ A P O in terms of the co-ordinates of point P and then taking the limit as P → O .
Let P ≡ ( t , 2 t ( t − 1 ) ) , as it lies on the parabola y = 2 x ( x − 1 ) .
Slope of O P , m 1 = t − 0 2 t ( t − 1 ) − 0 = 2 ( t − 1 ) .
Slope of P A , m 2 = t − 1 2 t ( t − 1 ) − 0 = 2 t .
Let ∠ A P O = θ , then ∣ tan θ ∣ = ∣ ∣ ∣ 1 + m 1 m 2 m 2 − m 1 ∣ ∣ ∣ .
Since sec 2 θ = 1 + tan 2 θ , we have sec θ = ± 1 + ( 1 + 4 t ( t − 1 ) 2 ) 2 .
Observe that as P → O , t → 0 ; and also ∠ A P O > 9 0 ∘ , so sec ∠ A P O < 0 .
P → O lim sec ∠ A P O = − t → 0 lim 1 + ( 1 + 4 t ( t − 1 ) 2 ) 2 = − 5 .
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There must be other ways to solve this, but I found this method quite traditional.
First of all,
Letting ∠ A P O = α ,
If sec α has a limit, then sin α also has a limit. Thus, we shall first find \displaystyle\sin\alpha).
Consider a general point P ( x , y ) , which lies on the given parabola.
It is clear that,
y = 2 x ( x − 1 )
Using Sine Rule,
O A sin α = A P sin ∠ A O P
Thus,
1 sin α = A P x 2 + y 2 y = ( x − 1 ) 2 + y 2 x 2 + y 2 y
Substituting y = 2 x ( x − 1 ) , we get,
sin α = ( 4 x 4 − 8 x 3 + 5 x 2 ) ( 4 x 4 − 8 x 3 + 5 x 2 − 2 x + 1 ) 2 x ( x − 1 ) = ( 4 x 2 − 8 x + 5 ) ( 4 x 4 − 8 x 3 + 5 x 2 − 2 x + 1 ) 2 ( x − 1 )
Note that we cancelled out the x 's, which is why we could not just substitute the final x-coordinates i.e x = 0
Now, substituting x = 0 , because, as the problem suggests, finally, x → 0
sin α = 5 − 2
⇒ cos α = ± 5 1
We reject the positive value, because, when P approaches O , clearly, ∠ A P O > 9 0 o
⇒ sec ( A P O ) < 0
Thus,
sec α = − 5