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
⇒ 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.
Prove that the circle with any focal chord of the parabola y2 = 4ax as its diameter touches its directrix.
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.
Find the locus of the centre of the circle described on any focal chord of aparabola as diameter.
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
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.
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 parabolabecomes y12 – 4ax1 > 0.
Similarly, the condition for P to lie inside the parabola is y12 – 4ax1 < 0.