Propositions on Parabola

Parabola is an extremely important topic of IIT JEE Mathematics syllabus. The standard equation of a parabola is y2 = 4ax and it opens sideways.The figure of parabola given below illustrates the focus, directirx, axis, vertex and various other components of a parabola.   

     parabola-tangent           

All these components have been already discussed in the previous seections. We proceed towards the various propositions on parabola:

  • The foot of the perpendicular from the focus on any tangent to a parabola lies on the tangent at the vertex.

Equation of the perpendicular to the tangent ty = x + at2 … (1)

From the focus (a, 0) is tx + y = at.                               … (2)

By adding (1) and (2) we get x = 0.(Since (1 + t2≠ 0)

Hence, the point of intersection of (1) and (2) lies on x =  0 i.e. on y-axis. This is also the tangent at the vertex of the parabola.

  • The tangent at any point P on a parabola bisects the angle between the focal chord through P and the perpendicular from P on the directrix.

The tangent at P (at2, 2at) is ty = x + at2.

It meets the x-axis at T(–at2, 0).

Hence, from the figure given above ST = SA + AT = a (1 + t2). 

Also, SP = √(a2(1 + t2)+ 4a2 t2 ) = a(1 + t2) = ST, so that 

∠MPT = ∠PTS = ∠SPT ⇒ TP bisects ∠SPM.

  • The portion of a tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus.

Let P(at2, 2a), be a point on the parabola y2 = 4ax.

Then the equation of tangent at P is ty = x + at2.

Point of intersection of the tangent with the directrix x + a = 0 is (–a, at – a/t).

Now, slope of SP is (2at-0)/(at2-a) = 2t/(t2-1) and slope of SK is (at-a/t-0)/(-a-a) = -(t2-1)/2t 

⇒ (Slope of the SP).(Slope of SK) = –1.

Hence SP is perpendicular to SK i.e. ∠KSP = 90°. 

  • Tangents at the extremities of any focal chord intersect at right angles on the directrix. 

Let P(at2, 2at) and P(at12, 2at1) be the end points of a focal part on the parabola. Then t.t1 = –1. Equations of the tangents at the point P and the point P’ are ty = x + at2 and t1y = x + at12 respectively.

Let these tangents intersects at a point (h, k). Then h = att1 and k = a(t + t1).

Since the tangents are perpendicular, tt1 = – 1 ⇒ h – a.

Hence the locus of the point (h, k) is x = –a which is the equation of the directrix. 

Some other Geometrical Properties are listed below:

  • The sub-tangent at any point on the parabola is twice the abcissa or proportional to square of the ordinate of the point.

  • The sub-normal is constant for all points on the parabola and is equal to its semi latus rectum 2a.

  • The locus of the point of intersection of the tangent at P and perpendicular from the locus on this tangent is the tangent at the vertex of the parabola.

  • In case a circle intersects a parabola in four points, the sum of the ordinates is always zero.

  • The area of the triangle formed by the three points on a parabola is twice the area of the traingle formed by the tangents at these points.

  • Tangents and normals at the extremities of the latus rectum of a parabola y2 = 4ax form a square and their points of intersection are (-a, 0) and (3a, 0).

  • The circle circumscribing the triangle formed by three tangents to a parabola always passes through the focus.

  • If the vertex and focus of parabola are on the x-axis and at a distance ‘a’ and ‘b’ from the origin respectively, then the equation of the parabola is y2 = 4(a’-a)(x-a).  

  • The orthocentre of any triangle formed by three tangents to a parabola y2 = 4ax lies on the directrix and has the coordinates as a and a(t1 + t2 + t3 + t1t2t3).

  • The area of the triangle formed inside the parabola y2 = 4ax is 1/8a (y1-y2)(y2-y3)(y3-y1) where y1, y2, y3 are the ordinates of vertices of the triangle.  

  • If the tangents at P and Q meet in T, then :

(a) TP and TQ subtend equal angles at the focus S.

ST2 = SP.SQ

The triangles SPT and STQ are similar.

 

We now discuss some of the illustrations based on parabola:

Illustration: 

The equation of the common tangent to the curves y2 = 8x and xy = -1 is (IIT JEE 2002)

(a) 3y = 9x + 2                                                               (b) y = 2x + 1

(c) 2y = x + 8                                                                 (d) y = x + 2

Solution: 

Tangent to the curve y2 = 8x is y = mx + 2/m.

Hence, it must satisfy xy = -1.

Hence, x(mx + 2/m) = -1

This gives mx2 + 2/m x + 1= 0 

Since, it has equal roots, so we must have D = 0

This gives 4/m2 – 4m = 0

so, m3 = 1 which gives m = 1.

Hence, the equation of the common tangent is y = x + 2.

Illustration: 

Normals are drawn from the point P with slopes m1, m2, m3 to the parabola y2 = 4x. If locus of P with m1m2 = a is a part of the parabola itself, then find a. (IIT JEE 2003)

Solution:

We know that the equation of the normal to  y2 = 4x is y = mx – 2am – am3

Thus, the equation of normal to  y2 = 4x is, y = mx – 2m – m3

Suppose that it passes through (h,k).

We have k = mh-2m-m3

or m3 + m(2-h) + k = 0

Here, m1 + m2 + m3 = 0

Also, m1m2 + m2m3 + m3m= 2-h

and m1m2m3 = -k, where m1m2 = a

This gives m3 = -k/a and this must satisfy (1)

This gives -k3/a3 – k/a(2-h) + k = 0

Hence, k2 = a2h – 2a2 + a3

Hence, y2 = a2x – 2a2 + a3

On comparing with y2 = 4x, we get

a2 = 4 and -2a2 + a3 = 0

This gives a =2.

Illustration:

At any point P on the parabola y2 – 2y – 4 x + 5 = 0, a tangent is drawn which meets the directrix at Q. Find the locus of point R which divides QP externally in the ratio 1/2:1. (IIT JEE 2004)

Solution:

We can rewrite the given equation as (y-1)2 = 4(x-1)

And its parametric coordinates are x-1 = t2 and y-1 = 2t

Hence, the coordinates of P are P(1+t2, 1+2t)

Equation of tangent at P is t(y-1) + x-1+t2

And this meets the directrix x = 0 at Q.

Thus, y = 1 + t-1/t or Q(0, 1 + t – 1/t)

Let us assume that K(h,k) is the point which divides QP externally in the ratio 1/2:1 or Q is the mid-point of KP.

Hence, 0 = (h + t2+1)/2 or t2 = – (h+1) ….... (1)

and 1+t-1/t = (k+2t+1)/2 or t = 2/(1-k) ….....(2)

Hence, from the above two equations we get

4/(1-k)2 + (h+1) = 0

or (k-1)2 (h+1) + 4 = 0

Hence, the locus of a point is (x+1)(y-1)2 + 4 = 0.

 

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