Focal Chord

Any chord to y² = 4ax which passes through the focus is called a focal chord of the parabola y² = 4ax.

Let y² = 4ax be the equation of a parabola and (at², 2at) a point P on it. Suppose the coordinates of the other extremity Q of the focal chord through P are (at12, 2at1).

Then, PS and SQ, where S is the focus (a, 0), have the same slopes

⇒ (2at-0)/(at²-a)=(2at₁-0)/(at₁²-a)

⇒ tt1² – t = t₁ t₂ – t1  (tt₁ + 1) (t₁ – t) = 0.

Hence t₁ = –1/t, i.e. the point Q is (a/t₂, –2a/t).

The extremities of a focal chord of the parabola y² = 4ax may be taken as the points t and –1/t.

Illustration:

Prove that the circle with any focal chord of the parabola y² = 4ax as its diameter touches its directrix.

Solution:

Let AB be a focal chord. If A is (at², 2at), then B is (a/t² ,-2a/t).

Equation of the circle with AB as diameter is

(x – at²) (x-a/t² ) + (y – 2at) (y+2a/t) = 0.

For x = –a, this gives (a² (1+t² )²)/t² + y² – 2ay (t-1/t) – 4a² = 0.

⇒ a² (t-1/t)² + y² – 2ay(t – 1/t) = 0

⇒ [y – a(t – 1/t)]² = 0, which has equal roots.

Hence x + a = 0 is a tangent to the circle with diameter AB.

Illustration:

Find the locus of the centre of the circle described on any focal chord of aparabola as diameter.

Solution:

Let the equation of the parabola be y² = 4ax.

Let t₁, t₂ be the extremities of the focal chord. Then t1 . t2 = – 1.

The equation of the circle on t1, t2 as diameter is

(x – at₂²) (x – at₂²) + (y – 2at₁) (y – 2at²) = 0

or x² + y² – ax (t₁² + t₂²) – 2ay (t₁ + t₂) + a² t₁2 t₁² + 4a² t₁t₂ = 0

⇒ x² + y² – ax (t₁² + t₂²) – 2ay (t₁ + t₂) – 3a² = 0. (∵ t₁t₂ = –1)

If (α,β) be the centre of the circle, then α = a/2 (t₁²+t₂² ) If (α, β) be the centre of the circle, then α = a/2 (t₁²+t₂² )

β = a (t₁ + t₂) ⇒ (t₁ + t₂)² =β²/a² ⇒ t₁² + t₂² + 2t₁t₂ =β²/a² ⇒ 2α/a-2= β²/a²

⇒ 2aα – 2a² = β² ⇒ β² = 2a (α – a).

Hence locus of (α, β) is y² = 2a(x – a).

Focal Distance of a Point

The focal distance of a point P on the parabola

parabola1

y² = 4ax is the distance between the point P and the focus S, i.e. PS. Thus thefocal distance of P = PS = PM = ZN = ZA + AN = a + x.

or

PS = a + at² = a(1 + t²).

Position of a point relative to a Parabola

Consider the parabola y² = 4ax.

If (x₁, y₁) is a given point and y₁² – 4ax₁ = 0, then the point lies on the parabola. But when y₁² – 4ax₁ ≠ 0, we draw the ordinate PM meeting the curve in L. Then P will lie outside the parabola if PM > LM, i.e., PM² – LM² > 0.

parabola2

Now, PM² = y₁² and LM2 = 4ax1 by virtue of the coordinates of L satisfying the equation of the parabola. Hence, the condition for P to lie outside the parabolabecomes y₁² – 4ax₁ > 0.

Similarly, the condition for P to lie inside the parabola is y₁² – 4ax₁ < 0.

Focal Chord

Focal Distance of a Point

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