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.   

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)² + 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 + m₃m₁ = 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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