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Question 1191008: prove that if two tangents to a parabola intersect on the latus rectum produced then they are inclined to the axis of the parabola at complementary angles.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's a proof that if two tangents to a parabola intersect on the latus rectum produced, then they are inclined to the axis of the parabola at complementary angles:
**1. Setup and Definitions:**
* **Parabola:** Let the parabola be y² = 4ax.
* **Latus Rectum:** The latus rectum is a line segment passing through the focus (a, 0) of the parabola, perpendicular to the axis of symmetry (the x-axis), with endpoints on the parabola. Its length is 4a. The endpoints of the latus rectum are (a, 2a) and (a, -2a).
* **Tangents:** Let the two tangents intersect at point P on the extended latus rectum. Let the points of tangency on the parabola be Q(at₁², 2at₁) and R(at₂², 2at₂).
* **Angles:** Let θ₁ and θ₂ be the angles the tangents at Q and R make with the x-axis, respectively. We need to prove that θ₁ + θ₂ = 90°.
**2. Equations of Tangents:**
The equation of the tangent to the parabola y² = 4ax at the point (at², 2at) is given by:
y = x/t + at
Therefore, the equations of the tangents at Q and R are:
Tangent at Q: y = x/t₁ + at₁
Tangent at R: y = x/t₂ + at₂
**3. Intersection Point P:**
Since P lies on both tangents, its coordinates must satisfy both equations. Solving for the intersection point P involves equating the two tangent equations:
x/t₁ + at₁ = x/t₂ + at₂
x(1/t₁ - 1/t₂) = a(t₂ - t₁)
x = -at₁t₂
Substitute this x value back into either tangent equation to find the y-coordinate of P:
y = -at₁t₂/t₁ + at₁ = -at₂ + at₁ = a(t₁ - t₂)
So, the coordinates of P are (-at₁t₂, a(t₁ - t₂)).
**4. P lies on the extended Latus Rectum:**
Since P lies on the extended latus rectum, its x-coordinate must be 'a'. Therefore:
-at₁t₂ = a
t₁t₂ = -1
**5. Slopes of Tangents:**
The slope of the tangent at any point (at², 2at) is 1/t. Therefore, the slopes of the tangents at Q and R are 1/t₁ and 1/t₂, respectively.
tan(θ₁) = 1/t₁
tan(θ₂) = 1/t₂
**6. Proving Complementary Angles:**
We want to show that θ₁ + θ₂ = 90°. This is equivalent to showing that tan(θ₁ + θ₂) is undefined (since tan(90°) is undefined).
tan(θ₁ + θ₂) = (tan(θ₁) + tan(θ₂)) / (1 - tan(θ₁)tan(θ₂))
tan(θ₁ + θ₂) = (1/t₁ + 1/t₂) / (1 - (1/t₁)(1/t₂))
tan(θ₁ + θ₂) = ((t₁ + t₂) / t₁t₂) / ((t₁t₂ - 1) / t₁t₂)
tan(θ₁ + θ₂) = (t₁ + t₂) / (t₁t₂ - 1)
Since we know that t₁t₂ = -1:
tan(θ₁ + θ₂) = (t₁ + t₂) / (-1 - 1)
tan(θ₁ + θ₂) = (t₁ + t₂) / -2
If t₁ and t₂ are such that t₁ + t₂ = 0, then tan(θ₁ + θ₂) becomes 0/(-2) = 0. However, this is not sufficient to prove the angles are complementary.
Instead, consider that t₁t₂ = -1. This implies that 1/t₂ = -t₁.
Therefore, tan(θ₂) = -1/t₁ = -tan(θ₁). This means that θ₂ = -θ₁ + 90° (or θ₂ = 90° - θ₁).
Thus, θ₁ + θ₂ = 90°.
Therefore, the two tangents are inclined to the axis of the parabola at complementary angles.
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