Properties of Focal Chord – I
Any point on the parabola y² = 4ax can be taken as (at², 2at), where t is parameter and t ϵ R. Any line passing through the focus of the parabola is called the focal chord of the parabola.
Example 1: If the chord joining P ≡ (at²₁, 2at₁) and Q ≡ (at²₂, 2at₂) is the focal chord, then t₁ t₂ = -1
Solution: P ≡ (at²₁, 2at₁) and Q ≡ (at²₂, 2at₂)
Since PQ passes through the focus S (a, 0), Q, S and P are collinear. Therefore,
Slope of PS = Slope of QS
\(\frac{2a{{t}_{1}}-0}{at_{1}^{2}-a}=\frac{0-2a{{t}_{2}}}{a-at_{2}^{2}}\).
\(\frac{2{{t}_{1}}}{t_{1}^{2}-1}=\frac{2{{t}_{2}}}{t_{2}^{2}-1}\).
t₁ (t₂²-1) = t₂ (t₁²-1)
t₁ t₂ (t₂ – t₁) + (t₂ – t₁) = 0
t₂ – t₁ ≠ 0
∴ t₁ t₂ + 1 = 0
t₁ t₂ = -1
t₂ = -1/t₁
Which is the required relation.
Note: If one extremity of a focal chord is (at²₁, 2at), then the other extremity (at²₂, 2at₂) becomes (a/t²₁, -2a/t₁).
Example 2: If point P is (at², 2at), then the length of the focal chord PQ is \(a{{\left( t+\frac{1}{t} \right)}^{2}}\).
Solution: PQ = SP + SQ
= a + at² + a + a/t²
= a (t + 1/t²)²
Example 3: The length of the focal chord which makes an angle θ with the positive direction of the x – axis is 4acosec²θ.
Solution: \(PQ=a{{\left( t+\frac{1}{t} \right)}^{2}}\).
Now, Slope of PQ = 2/ [t – (1/t)] = tanθ
2 cotθ = t – (1/t).
\(PQ=a{{\left( t+\frac{1}{t} \right)}^{2}}=a\left[ {{\left( t-\frac{1}{t} \right)}^{2}}+4 \right]\).
= a [4cot²θ + 4] = 4a cosec²θ
From this, we can conclude that the minimum length of the focal chord is 4a, which is the length of the latus rectum.