# Parametric Coordinates of Parabola , Focal Chord

Any point on the parabola y2 = 4ax is (at2, 2at) and we refer to it as the point ‘ t ‘ . Here, t is a parameter , i.e. , it varies from point to point.

Illustration : Show that the locus of a point that divides a chord of slope 2 of the parabola y2 = 4x internally in the ratio 1 : 2 is a parabola. Find the vertex of this parabola.

Solution: Let the two points on the given parabola be (t12, 2t1) and (t22, 2t2). Slope of the line joining these points is

$\large 2 = \frac{2t_2 – 2t_1}{t_2^2 – t_1^2} = \frac{2}{t_2 + t_1}$

⇒ t1 + t2 = 1

Hence the two points become (t12 , 2t1) and ((1 − t1)2 , 2(1 − t1))

Let (h , k) be the point which divides these points in the ratio 1 : 2

$\large h = \frac{(1-t_1)^2 + 2 t_1^2}{3} = \frac{1-2t_1 + 3t_1^2}{3}$ ….(1)

$\large k = \frac{2(1-t_1) + 4 t_1}{3} = \frac{2 + 2t_1 }{3}$ ….(2)

Eliminating t1 from (1) and (2), we find that 4h = 9k2 − 16k + 8

Hence locus of (h , k) is

$\large (y-\frac{8}{9})^2 = \frac{4}{9}(x-\frac{2}{9})$

This is a parabola with vertex (2/9 , 8/9)

##### 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.

$\large \frac{2at-0}{at^2-a}= \frac{2at_1-0}{at_1^2-a}$

⇒ t t12 − t = t1 t2 − t1

⇒ ( t t1 + 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 : Through the vertex O of a parabola y2 = 4x chords OP and OQ are drawn at right angles to one another. Show that for all position of P, PQ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of PQ.

Solution: The given parabola is y2 = 4x ….(1)

Let P ≡ (t12, 2t1) , Q ≡ ( t22 , 2t2 )

Slope of OP = 2t1/t12 = 2/t1 and slope of OQ = 2/t2

Since OP ⊥ OQ ,

4/t1t2 = −1 or t1t2 = −4  ….(2)

The equation of PQ is

$\large y-2t_1 = \frac{2(t_1-t_2)}{t_1^2-t_2^2}(x-t_1^2)$

Or y ( t1 + t2) = 2 ( x + t1t2)

⇒ y ( t1 + t2) = 2 (x − 4) [ from (2)]

This line cuts the x-axis (the axis of the parabola) at (4, 0) which is a fixed point for all positions of P

Let R(α, β) be the middle point of PQ. Then

$\large \alpha = \frac{t_1^2 + t_2^2}{2}$ …(3)

and β = t1 + t2 …(4)

From (4), β2 = t12 + t22 + 2t1 t2 = 2α – β [ From (2) and (3)]

Hence locus of R(α , β) is y2 = 2x − 8

Illustration : A tangent and a normal is drawn at the point P≡ (16 , 16) of the parabola y2 = 16x which cut the axis of the parabola at the points A and B respectively. If the centre of the circle through P, A and B is C then find the angle between PC and the axis of x.

Solution:

By property centre of circle coincides with focus of parabola

⇒ C ≡ (4 , 0)

tan α = slope of PC = 16/12

⇒ α = tan−1(4/3)

Focal Distance of any point

The focal distance of any 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 = PS = PM

ZN = ZA + AN = a + x

###### Position of a point relative to the 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.

Substituting these values in equation of parabola, 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 : For what values of ‘α’ the point P(α, α) lies inside, on or outside the parabola (y − 2)2 = 4(x − 3)

Solution: Given equation can be written as y2 − 4y − 4x + 16 = 0

Point P(α , α) lies inside parabola if α2 − 8α + 16 < 0

⇒ (α − 4)2 < 0

⇒ no such α exist.

Point P(α, α) lies on parabola if (α − 4)2 = 0

⇒ α = 4

Point P(α, α) lies outside parabola if (α − 4)2 > 0

⇒ α ∈ R − {4}

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