E-StoreHome » Ask & Discuss » Mathematics. » Analytical Geometry « Back to Discussion
Analytical Geometry
Comments (25)
let the foot of perpendicular drawn from origin O(0,0) on the chord AB of the given circle be P(h,k) such that the chord AB substends a right angle at origin
eq, of chord AB is .................y-k=-h/k(x-h)
hx + ky = h^2+ k^2
___________________________________________________________________________________
________________________________________________________________________________
the combined eq of OA and OB is homogeneous EQ of SECOND DEGREE which is obtained by the help of given circle and the chord AB ....IS GIVEN BY ......x^2 + y^2 +(2gx +2fy)(hx +ky)/(h^2 + k^2) + c((hx +ky)/(h^2 + k^2))^2 =0...............(i)
since the lines OB and OA given by above eq are at right angle ....hence
coefficient of x^2 + coefficient of y^2 =0
on putting the coefficient of x^2 + coefficient of y^2 =0 from eq (i) eq of locus comes out to be x^2 + y^2 + gx + fy +c/2 =0.......
________________________________________________________________________________
for Q2. let the point be P(at2,2at). The eqn of tangent is ty=x+at2 eqn of normal is y= -tx + at3 + 2at ....
equation of line joining focus (a,0) and P is (t2-1)y = 2tx -2at
angle between this curve and tangent is tan-1(2t/(t2-1) - 1/t)/(1+2t/(t2-1)*1/t) = tan-1(1/t)
The equation of line perpendicular from P on the directrix has slope zero.
So angle between this curve and tangent is tan-1(1/t)/(1-0) = tan-1(1/t)
As the angles are equal. So the tangent bisects the angle between the two given lines (and as the normal is perpendicular to tangent it will bisect the other pair of angles)
First i m proving for tangents Let P be point t .
PS = PM = a +a t2
let the tangent meet x- axis at Q
so coordinates of Q is ( -at2 , 0 )
SQ = a +a t^2
so angle PQ S= angle Q P S
but angle PQS= angle MPS
thus PQ is bisector of MPS
Proving it for normals - i don't think any one need my help to prove this part too .
For a parabola y2 = 4ax
equation of normal is given by....................................... y = mx-2am-am3
Ok..
this one is a big one...After this we'll have objectives.
Question 4 :
From a point (h,k) three normals are drawn to the parabola y^2 = 4ax.
Now a circle passing through the point of intersection of the normals meet the parabola at four points.
Find the co-ordinates of the fouth point and equation of circle.
Answer 4:
Normal of slope m ---> y = mx - 2am - am^3
As it passes through (h,k),
k = mh - 2am - am^3
Instead, let us follow the approach of normal at a point as we have to deal with points.
Normal at P (at^2, 2at) ---> y + tx = 2at + at^3
As (h,k) lies on it,
k + th = 2at + at^3 ----> at^3 + (2a-h)t - k = 0 ......................(A)
So the three points given by t1, t2, t3 where t1 + t2 + t3 = 0 .............(1)
Let circle by x^2 + y^2 + 2gx + 2fy + c = 0
As it also intersects the parabola at P (at^2, 2at)
a^2.t^4 + 4a^2.t^2 + 2agt^2 + 4aft + c = 0
-----> a^2.t^4 + (4a^2 + 2ag)t^2 + 4aft + c = 0 ....................(B)
It intersects the parabola at four points t1, t2, t3, t4.
t1+ t2 + t3 + t4 = 0 .....................(2)
From (1) and (2)
t4 = 0 or fourth point of intersection is origin.
Also, summation ( t1t2) = (4a^2 + 2ag) / a^2 = 4 + 2g/a
t1t2 + t2t3 + t1t3 + t4(t1 + t2 + t3) = 4 + 2g/a
(written in this order as t4 = 0)
t1t2 + t2t3 + t3t1 = 4 + 2g/a
From (A)
t1t2 + t2t3 + t3t1 = (2a-h) / a = 2 - h/a
So, 4 + 2g/a = 2 - h/a
2g = -a ( h/a + 2) = -h - 2a
g = -h/2 - a
From (B),
t1.t2.t3 + t4 (t1t2 + t2t3 + t3t1) = -4af / a^2 = -4f / a
hence, t1.t2.t3 = -4f / a
From (A)
t1.t2.t3 = + k/a
hence, f = -k/4
From (B) .. c = t1.t2.t3.t4 = 0
So circle is
x^2 + y^2 - (h + 2a) x - ky / 2 = 0
And as already proved above ..the fourth point is origin.
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
Find Posts by Topics
Physics.
TopicsMathematics.
Chemistry.
Biology
Institutes
Parents
Board
Fun Zone
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
1951 Students Currently Online |
Connect with Us  |
 |
Question 1 :
A variable chord of the circle x^2 + y^2 + 2gx + 2fy + c = 0 always subtends a right angle at the origin.
Find the LOCUS of the foot of the perpendicular drawn from origin to this chord.
For answer, try selecting text..
< The locus of P is :
2( x^2 + y^2 )+ 2gx + 2fy + c = 0 >
< One of the best basic questions on homogenisation.>