IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

Judasrising

i need to know the principle behind homogenising 2 or more curves. I understand that a homogenous curve gas a fixed degree in each of its terms but the concept of "homogenising" curves is not understood by me.

for ex. consider a parabola : y²= 4ax,

say a chord PQ (whose mid-pt. is [h,k]) subtends an angle of 90 degrees at the vertex. (A).The eqn. of the chord PQ is,

ky-2a(h+x)=k² - 4ah .........(1)

the combined eqn of the lines AP and AQ is obtained by "homogenising" the eqn. of PQ with the parabola.

the process used is,

since, y²=4ax,

y²=4ax [(yk-2ax)/(k²-2ah)]................(since [(yk-2ax)/(k²-2ah)]=1)

(i saw this ex. in Arihant's co-ord geom. by SK goyal)

on simplifying this, we get the desired eqns..

my question is, WHAT IS "HOMOGENISING" HOW DO WE KNOW THAT "HOMOGENISING" GIVES US THIS EQUN. HOW DO WE HOMOGENISE 2 GENERAL CURVES. WHAT IS THE INTERPRETATION OF THE RESULT OF THIS PROCESS?

please reply.

eistien

my interpretation of it is that any 2 degree equation can be expressed as a product of 2 1degree equations similarly a curve can be expressed as 2 straight lines wrt to a point on the plane emerging from that part point which makes various problems much simpler for solving as you would have seen in arihant book

iitjee08aspirant

i have same doubt ....

puneet

hi

well .. I do not knw the gr8 details of this stuff .. but i shall try to give a overview .. as to how this works ..

the main idea is that if we hav two curves C and D .. thn any curve passing through the two curves shall take the form C + k.D .. where k is some constant ..

another concept is from the pair of straight lines .. which say tat two lines can be written as product of one another and thus resulting in the quadratic equations ..

this two are the reasons y we do the curve homogenization .. i shall try to look for more stuff .. and shall get to u if i find something more interesting for u ..

i must appreciate u for raising this point .. u hav a bright future ahead :)

well done .

cheers

bvsatyaram

Let ax2.by2+............+c=0, represent the equation of a general conic section.

Let mx+ny+p=0 be the equation of a chord. Now, as pointed out by Judasrising, the homogeneousing technique results in a pair of straight lines, with origin as their point of intersection and this pair passes through the point of intersection of chord and the conic section.

Lets now look at the justification part:

- The resulting equation is homogeneous. So, (0,0) is a point on the resulting homogeneous equation.

- Let (x1,y1) and (x2,y2) be the points of intersection of the chord and the conic section. It can be easily seen that these two points also satisfy the homogeneous equation.

So, now we conclude that the resulted homogeneous equation is the equation of a curve, passing through origin and 2 points of intersection of the chord and the conic section. The only which we need to prove is that the equation represents a pair of straight lines.

Try getting a condition for the homogeneous equation to be pair of straight lines. the condition we get will be same as the condition for the chord to intersect the conic section. So, the homogeneous equation represents a pair of straight lines, when ever the given straight line intersects with the conic section.

Hence justified.

konichiwa2x

If the line

for

i.e the line not passing though origin cuts the curve

at two points A and B, then the joint equation of the straight lines passing though A and B and the origin is given by the homogenising the equation of the curve by the equation of the line. To homegenise, consider the curve with the higher defree. And then multiply it by the first curve WITHOUT changing it.

i.e

So we now multiply with this quantity such that every term is of same power.

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