Focal Chord of Parabola

Intersection of a Straight Line with a Parabola

The combined equation of straight line y = mx + c and parabola y² = 4ax gives us the co-ordinates of point(s) of their intersection.

The combined equation m²x² + 2x (mc – 2a) + c² = 0 will give those roots. The straight line therefore meets the parabola at two points. 

Points of intersection of y² = 4ax and y = mx + c are given by (mx+c)² = 4ax i.e. m²x² + 2x(mc – 2a) + c² = 0. …… (i) 

Since (i) is a quadratic equation, the straight line meets the parabola in two points, real, coincident, or imaginary. The roots of (i) are real or imaginary according as {2(mx – 2a)}² – 4m²c² is positive or negative, i.e. according as – amc + a² is positive or negative, that is, according as mc is less than or greater than a, (taking a as positive). 

Note: 

When m is very small, one of the roots of equation (i) is very large; when m is equal to zero, this root is infinitely large. Hence every straight line parallel to the axis of the parabola meets the curve in one point at a finite distance and in another point at an infinite distance from the vertex. It means that a line parallel to the axis of the parabola meets the parabola only in one point. 
 

Chord with a given middle point

Equation of the chord of the parabola y² = 4ax whose middle point is (x₁, y₁) is 

(y-y₁) = 2a/y₁(x-x₁)

This can be written as T = S₁, where T = yy₁ – 2a(x+x₁) and S₁ = y₁² – 4ax₁.
 

Focal Chord of a Parabola

The chord of the parabola which passes through the focus is called the 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 = t₁t² – 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. 
 

Length of the chord 

The abscissae of the points common to the straight line y = mx + c and the parabola y² = 4ax are given by the equation m²x² + (2mx – 4a) x + c² = 0. 

= 4(mc-2a)² / m⁴ – 4c²/m² = 16a(a-mc)/ m⁴ and (y₁ – y₂) = m(x₁-x₂)

Focal Distance of a Point 

The focal distance of a point P on the parabola

y² = 4ax is the distance between the point P and the focus S, i.e. PS. Thus the focal 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. 
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 parabola becomes y₁² – 4ax₁ > 0. 

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

Illustration: 

Find the Length of the chord intercepted by the parabola y² = 4ax from the line y = mx + c. Also find its mid-point.

Solution: 

Simply by applying the formula of length of the line joining (x₁, y₁) and (x₂, y₂) we get, 

Length of the chord = √((x₁-x₂)² + (y₁-y₂)² ) 

= √((x₁-x₂)² + m²(x₁-x₂)²) 

= |x₁ – x₂|√(1+m² ) = 4√(a(a-mc)) √(1+m² ) 

[ ? x₁ + x₂ = (-2(m-2a))/m² and x₁x₂ = c²/m² ] 

The midpoint of the chord is ((2a-mc)/m², 2a/m)

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 a parabola 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). 
 

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