Home » Ask & Discuss » Mathematics. » Analytical Geometry « Back to Discussion

Analytical Geometry

Cool goIITian

Joined:	9 Sep 2007
Post:	39

22 Oct 2007 15:27:31 IST

0 People liked this

300

ANSWER THIS PLZ

None

A VARIABLE LINE DRAWN THROUGH THE POINT OF INTERSECTION OF THE ST.LINES X/

+ Y/

=1 & X/

+Y/

=1 MEET THE COORDINATE AXIS AT A AND B .IF M IS MIDPOINT OF AB SHOW THAT LOCUS OF M IS THE CURVE 2(

)XY=

(X+Y).

Share this article on:

Comments (1)

Bipin Dubey

Forum Expert

Joined: 23 Jan 2007

Posts: 7942

22 Oct 2007 18:17:34 IST

Like 1 people liked this

I am taking a and b in place of alpha and beta for convenience.

Lines are : x/a + y/b = 1 and x/b + y/a = 1

Solve these two equations with two variables x and y to find the intersection point. You will get x = ab/(a+b) and y = ab/(a+b).

Any variable passing through this point can be expressed as :

y - ab/(a+b) = m{x - ab/(a+b)}

It intersects x-axis at A where y=0, A = { (ab/a+b)(1 - 1/m) , 0 }

It intersects y-axis at B where x=0, B = { 0 , (ab/a+b)(1 - m) }

Mid-point of AB = { (ab/a+b)(1 - 1/m)/2 , (ab/a+b)(1 - m)/2 }

Let it be (h.k)

h = (ab/a+b)(1 - 1/m)/2 and k = (ab/a+b)(1 - m)/2

Dividing these two we get h/k = -1/m

Substituting this in the expression of h we get :

h = (ab/a+b)(1 + h/k)/2

Rearranging this we get :

1/h + 1/k = 2(1/a + 1/b)

Hence locus of (h,k) is

1/x + 1/y = 2(1/a + 1/b)

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team