Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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.
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:
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
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.
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)
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.
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)
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.
Click here to refer the most Useful Books of Mathematics.
For getting an idea of the type of questions asked, refer the previous year papers.
To read more, Buy study materials of Parabola comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Chord of a Parabola Intersection of a Straight...
Focal Chord of Parabola Table of Content...
Normal to a Parabola What do we mean by a normal...
Tangent to a Parabola Table of Content Equation of...
Introduction to Parabola Conics or conic sections...
Solved Examples on Parabola Parabola is an...