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).


                            parabola
        

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.
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