**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.