Proving ellipse property (1 Viewer)

[ashokkumar]

Show that the sum of the distances from any point P on the ellipse x^2/a^2 + y^2/b^2 = 1 to the two foci is equal to 2a.

 

[thongetsu]

penis

 

[nat_doc]

us the fact that PS=ePM.... hence the distance of the point is

PS+PS' = e|PM+PM'|
let the point be (x,y)
the sum of the distances from the focii to the point is
PS+PS' =e|distance of P to M and P from M'|=e|(a/e-x)+(a/e +x)|=e|2a/e|=2a
NOTE: the distances can be easily seen from a diagram.

 

[ashokkumar]

ah is that short method allowed? Because the method using distance formula is long and confusing at the end.

why is the absolute of (ex - a) = a - ex?

 

[omgmynameislong]

|5-2|=|2-5|

 

[ashokkumar]

|5-2|=|2-5|

but why not make the absolute value of (ex - a) = ex - a, instad of a - ex?

 

[Timothy.Siu]

well remember distance/absolute value is always positive.
does a-ex>0 for all x, for ellipses then?

 

[seanieg89]

yes, because x=a/e is the equation of the right directrix.

 

[nat_doc]

ah is that short method allowed? Because the method using distance formula is long and confusing at the end.

why is the absolute of (ex - a) = a - ex?

|ex-a|= |-1||a-ex|= 1*|a-ex|