Conics Question (1 Viewer)

[Harimau]

(From page 77 of fitzpatricks)

Using Cartesian coordinates, show that the absolute value of the differewnce of the distances from any point P on the hyperbola x^2/a^2 + y^2/b^2 =1 to the two foci is equal to 2a.

Is there a "Quick" way of doing this rather than the 1-2 pages of algebra?

 

[spice girl]

Originally posted by Harimau
(From page 77 of fitzpatricks)

Using Cartesian coordinates, show that the absolute value of the differewnce of the distances from any point P on the hyperbola x^2/a^2 + y^2/b^2 =1 to the two foci is equal to 2a.

Is there a "Quick" way of doing this rather than the 1-2 pages of algebra?

btw, it's an ellipse u have there. the hyperbola should be x^2/a^2 - y^2/b^2 =1

but, yes.

hyperbola: alternate definition: the absolute value of the difference of lengths to two set points is a constant.

i.e. ||SP| - |S'P|| = k

and |SP| = e|PM| where e is eccentricity, M is the base of perpendicular dropped from P to the directrix.

we know the directrices are equation x = +- a/e, so MM' (M' is basically M, but to the OTHER directrix) = a/e - (-a/e) = 2a/e

so ||SP|-|SP'|| = e||PM| - |PM'|| = e|MM'| (drawing this out will make it more obvious)

e|MM'| = e(2a/e) = 2a

 

[Harimau]

Re: Re: Conics Question

Originally posted by spice girl


btw, it's an ellipse u have there. the hyperbola should be x^2/a^2 - y^2/b^2 =1

but, yes.

hyperbola: alternate definition: the absolute value of the difference of lengths to two set points is a constant.

i.e. ||SP| - |S'P|| = k

and |SP| = e|PM| where e is eccentricity, M is the base of perpendicular dropped from P to the directrix.

we know the directrices are equation x = +- a/e, so MM' (M' is basically M, but to the OTHER directrix) = a/e - (-a/e) = 2a/e

so ||SP|-|SP'|| = e||PM| - |PM'|| = e|MM'| (drawing this out will make it more obvious)

e|MM'| = e(2a/e) = 2a

I dont quite understand this step...

we know the directrices are equation x = +- a/e, so MM' (M' is basically M, but to the OTHER directrix) = a/e - (-a/e) = 2a/e

 

[spice girl]

well if P (the locus pt) had the co-ords (x,y)

then M would have the co-ords (a/e, y)
because PM is a horizontal line, and M lies on the directrix x = a/e

and M' lies on the other directrix x = -a/e
so its co-ords are (-a/e, y)

so the distance MM' is obviously = 2a/e

 

[Harimau]

Originally posted by spice girl
well if P (the locus pt) had the co-ords (x,y)

then M would have the co-ords (a/e, y)
because PM is a horizontal line, and M lies on the directrix x = a/e

and M' lies on the other directrix x = -a/e
so its co-ords are (-a/e, y)

so the distance MM' is obviously = 2a/e

ooooh right. thank you!!!