IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

allamraju

It is simply based on the formula (a+b)(a-b)=a²-b².Let the first eqn. be p+q=2a.Then,(do this in mind) (p+q)(p-q)=(x-ae)²-(x+ae)²=-4aex $\Rightarrow$ 2a(p-q)=-4aex $\Rightarrow$ (p-q)=-2ex.This is what written in the 3rd eqn.Hope u got it.

studyid

Or else another approach :

by the definition of eccentricity ,

we have $\frac{AS}{AC} = e$ (see the figure)

hence ,

by distance formula , we have ,

$\frac{(x+ae)^{2}+{y}^{2}}{(x+\frac{a}{e})^{2}} = {e}^{2}$

hence ${x}^{2}+{y}^{2}+{a}^{2}{e}^{2}+2aex={e}^{2}{x}^{2}+{a}^{2}+2aex$

hence ${x}^{2}(1-{e}^{2})+{y}^{2}={a}^{2}(1-{e}^{2})$

hence $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}(1-{e}^{2})} =1$

Hence $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}} =1\:\: ...... eqn\:of\:ellipse$

FORCE_IIT

((x+ae)²+y²)^1/2+((x-ae)^1/2+y²)^1/2=2a

solve and you get the equation of an ellipse

www.bits360.com

BITS 1999 ???
Most likely a mistake.

rachit12

bhaiyo ml khanna main likha hain !!
kya karun/?

mukundmadhav

Dude this is definition of ellipse.. You should know this.. Ye sab to basic stuff hota hai.. You shouldn't start solving problems before learning all this.

www.bits360.com

Yes , thats true.
And besides its a basic question on locus.

The gen. strategy for such qns is to use the definition of that situation.Here the question talks about distances and them being equal.
So its natural to use distance formula on a variable point(h,k) , twice.
And then equate both , replace h>x , k>y .
Rearrange them to get it in the above form.

Sometimes a qn may ask a point which follows more constraints.Simply list out all the constraints , write them in mathematical form and try to create 1 eqn that links all of them.

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It is simply based on the formula (a+b)(a-b)=a2-b2.Let the first eqn. be p+q=2a.Then,(do this in mind) (p+q)(p-q)=(x-ae)2-(x+ae)2=-4aex2a(p-q)=-4aex(p-q)=-2ex.This is what written in the 3rd eqn.Hope u got it.

It is simply based on the formula (a+b)(a-b)=a²-b².Let the first eqn. be p+q=2a.Then,(do this in mind) (p+q)(p-q)=(x-ae)²-(x+ae)²=-4aex $\Rightarrow$ 2a(p-q)=-4aex $\Rightarrow$ (p-q)=-2ex.This is what written in the 3rd eqn.Hope u got it.