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, i.e. 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 y= 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).

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

⇒ (2at-0)/(at2- a) = (2at- 0)/(at1- 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

                                 parabola1

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.

                                 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 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)+ (y1-y2)2 ) 

= √((x1-x2)+ m2(x1-x2)2

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

[ ∵ x+ x= (-2(m-2a))/m2 and x1x= 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 (t+ 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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