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Propositions on Parabola Parabola is an extremely important topic of IIT JEE Mathematics syllabus. The standard equation of a parabola is y^{2} = 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:
Parabola is an extremely important topic of IIT JEE Mathematics syllabus. The standard equation of a parabola is y^{2} = 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:
Equation of the perpendicular to the tangent ty = x + at^{2} … (1)
From the focus (a, 0) is tx + y = at. … (2)
By adding (1) and (2) we get x = 0.(Since (1 + t^{2}) ≠ 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 P (at^{2}, 2at) is ty = x + at^{2}.
It meets the x-axis at T(–at^{2}, 0).
Hence, from the figure given above ST = SA + AT = a (1 + t^{2}).
Also, SP = √(a^{2}(1 + t^{2})^{2 }+ 4a^{2} t^{2} ) = a(1 + t^{2}) = ST, so that
∠MPT = ∠PTS = ∠SPT ⇒ TP bisects ∠SPM.
Let P(at^{2}, 2a), be a point on the parabola y^{2} = 4ax.
Then the equation of tangent at P is ty = x + at^{2}.
Point of intersection of the tangent with the directrix x + a = 0 is (–a, at – a/t).
Now, slope of SP is (2at-0)/(at^{2}-a) = 2t/(t^{2}-1) and slope of SK is (at-a/t-0)/(-a-a) = -(t^{2}-1)/2t
⇒ (Slope of the SP).(Slope of SK) = –1.
Hence SP is perpendicular to SK i.e. ∠KSP = 90°.
Let P(at^{2}, 2at) and P(at_{1}^{2}, 2at_{1}) be the end points of a focal part on the parabola. Then t.t_{1} = –1. Equations of the tangents at the point P and the point P’ are ty = x + at^{2} and t_{1}y = x + at_{1}^{2} respectively.
Let these tangents intersects at a point (h, k). Then h = att_{1} and k = a(t + t_{1}).
Since the tangents are perpendicular, tt_{1} = – 1 ⇒ h – a.
Hence the locus of the point (h, k) is x = –a which is the equation of the directrix.
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 y^{2} = 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 y^{2} = 4(a’-a)(x-a).
The orthocentre of any triangle formed by three tangents to a parabola y^{2} = 4ax lies on the directrix and has the coordinates as a and a(t_{1} + t_{2} + t_{3} + t_{1}t_{2}t_{3}).
The area of the triangle formed inside the parabola y^{2} = 4ax is 1/8a (y_{1}-y_{2})(y_{2}-y_{3})(y_{3}-y_{1}) where y_{1}, y_{2}, y_{3} 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.
ST^{2} = SP.SQ
The triangles SPT and STQ are similar.
We now discuss some of the illustrations based on parabola:
The equation of the common tangent to the curves y^{2} = 8x and xy = -1 is (IIT JEE 2002)
(a) 3y = 9x + 2 (b) y = 2x + 1
(c) 2y = x + 8 (d) y = x + 2
Tangent to the curve y^{2} = 8x is y = mx + 2/m.
Hence, it must satisfy xy = -1.
Hence, x(mx + 2/m) = -1
This gives mx^{2} + 2/m x + 1= 0
Since, it has equal roots, so we must have D = 0
This gives 4/m^{2} – 4m = 0
so, m^{3} = 1 which gives m = 1.
Hence, the equation of the common tangent is y = x + 2.
Normals are drawn from the point P with slopes m_{1}, m_{2}, m_{3} to the parabola y^{2} = 4x. If locus of P with m_{1}m_{2} = a is a part of the parabola itself, then find a. (IIT JEE 2003)
We know that the equation of the normal to y^{2} = 4x is y = mx – 2am – am^{3}
Thus, the equation of normal to y^{2} = 4x is, y = mx – 2m – m^{3}
Suppose that it passes through (h,k).
We have k = mh-2m-m^{3}
or m^{3} + m(2-h) + k = 0
Here, m_{1} + m_{2} + m_{3} = 0
Also, m_{1}m_{2} + m_{2}m_{3} + m_{3}m_{1 }= 2-h
and m_{1}m_{2}m_{3 }= -k, where m_{1}m_{2} = a
This gives m_{3} = -k/a and this must satisfy (1)
This gives -k^{3}/a^{3} – k/a(2-h) + k = 0
Hence, k^{2} = a^{2}h – 2a^{2} + a^{3}
Hence, y^{2} = a^{2}x – 2a^{2} + a^{3}
On comparing with y^{2} = 4x, we get
a^{2} = 4 and -2a^{2} + a^{3} = 0
This gives a =2.
At any point P on the parabola y^{2} – 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)
We can rewrite the given equation as (y-1)^{2} = 4(x-1)
And its parametric coordinates are x-1 = t^{2} and y-1 = 2t
Hence, the coordinates of P are P(1+t^{2}, 1+2t)
Equation of tangent at P is t(y-1) + x-1+t^{2}
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 + t^{2}+1)/2 or t^{2} = – (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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