perplexus.info :: Geometry : Orthogonal Tangents

Orthogonal Tangents (Posted on 2011-11-30)

Let Γ be a parabola with focus F and directrix d.
A line through F intersects Γ in points P₁ and P₂.
The feet of the perpendiculars from P₁ and P₂ on
d are Q₁ and Q₂ respectively.
The midpoint of line segment Q₁Q₂ is Q₀.

Prove that the rays Q₀P₁ and Q₀P₂ are orthogonal
and that they are tangent to Γ.

Submitted by Bractals

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Solution: (Hide)

Construct line segments FQ₁, FQ₂, and FZ where
Z lies on d and FZ ⊥ d. FQ₁ intersects Q₀P₁ at
point R₁ and FQ₂ intersects Q₀P₂ at point R₂

Since |P₁F| = |P₁Q₁|, ∠P₁FQ₁ = ∠P₁Q₁F = α.
FQ₁ is a transversal of parallel lines FZ and
P₁Q₁. Therefore, ∠Q₁FZ = ∠P₁Q₁F = α. Similarly,
∠P₂FQ₂ = ∠Q₂FZ = β.

Therefore, straight ∠P₁FP₂ = ∠P₁FQ₁ + ∠Q₁FZ +
∠P₂FQ₂ + ∠Q₂FZ = 2α + 2β. Thus,
∠Q₁FQ₂ = α + β = 90°.

Construct line segment FQ₀. Since Q₀ is the
midpoint of the hypotenuse of right triangle
Q₁FQ₂, |Q₀F| = |Q₀Q₁| = |Q₀Q₂|.

Therefore, P₁FQ₀Q₁ is a kite and the diagonals
of a kite are perpendicular. Similarly, the
diagonals of P₂FQ₀Q₂ are perpendicular.

Three of the interior angles of quadrilateral
FR₁Q₀R₂ are right angles. Thus, the fourth
∠R₁Q₀R₂ is also.

Thus, the rays Q₀P₁ and Q₀P₂ are orthogonal.

QED

Because P₁FQ₀Q₁ and P₂FQ₀Q₂ are kites,
the diagonals Q₀P₁ and Q₀P₂ bisect ∠FP₁Q₁ and
∠FP₂Q₂ respectively.

Thus, the rays Q₀P₁ and Q₀P₂ are tangent to Γ.

QED

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: solution	Bractals	2011-12-01 17:11:27
solution	Daniel	2011-12-01 10:35:19

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