Question

Let y² = 4ax be a parabola with focus S(a, 0) and directrix x + a = 0. Then, the following properties hold good.

1.  If P (at₁² , 2at₁) is one end of the focal chord, then the other end is

(a/t₁², -2a/t₁)

That is, if t is the parameter of one end of the focal chord, then the parameter of the other end is –1/t.

2.  Length of the focal chord whose one end is (at², 2at) is

a(t + 1/t)²

3.  If PSQ is focal chord, then

1/SP + 1/SQ = 1/a = 2/2a

In other words, the semi-latus rectum is the harmonic mean between the focal radii of the ends of a focal chord.

4. The circle described on a focal chord touches the directrix.

5. The circle described on a focal radius of a point on the parabola touches the tangent at the vertex. 

6. The circle described on a focal radius of a point P(at², 2at) cuts an intercept of length a√1 + t² on the normal at P. 

7. The tangents drawn at the extremities of a focal chord are at right angles and hence intersect on the directrix of the parabola.

Answer 1

1.  Let Q (at₂² , 2at₂) be the other end of the focal chord for which P (at₁² , 2at₁) is one end. Equation of the focal chord PQ is

This chord passes through S(a, 0). So we have

2.  Let P(at², 2at) be one end of the focal chord so that by point (1), the other end is

3.  Let P = (at², 2at) and Q = (a/t² ,-2a/t) . Then, we know that

and 2a is the semi-latus rectum. Hence, SP, semi-latus rectum, and SQ are in HP.

4.  P(at², 2at) and Q(a/t², - 2a/t) are the ends of a focal chord. It is known that the centre of the circle described on PQ as diameter is

 

and radius [by point (2)] is

Now, the distance of the centre of the circle from the directrix is

Hence, the circle described on PQ as diameter touches the directrix.

5. The circle with SP as diameter is (x - a)(x - at²) + y(y - 2at) = 0. The centre is

and the radius is

Distance of the centre from the y-axis is the radius, which is given by

a/2(1 + t²)

Thus, the circle described on SP as diameter touches the tangent at the vertex.

6. Let the circle described on SP meet the normal at P in N. Draw SM perpendicular to the tangent at P. Since SNPM is a rectangle, we have (see Fig.)

7.  Let P(at², 2at) and Q(a/t² , -2a/t) The tangent at P is

ty = x + at²

whose slope is 1/t. The tangent at Q is

-y/t = x + a/t²

whose slope is –t. Therefore, the tangents at P and Q intersect at right angles.