• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

ellipse question (1 Viewer)

c0okies

Member
Joined
Nov 12, 2005
Messages
132
Location
here
Gender
Female
HSC
2006
The equation of the tangent of p(x1,y1) on the ellipse E:

x^2/a^2 + y^2/b^2=1 is xx1/a^2 + yy1/b^2 =1

It meets the major axis in T and a directrix in Q, PN is the ordinate of P, S is the focus corresponding to this directrix. Show that:
a) ON.OT=a^2 (done)
b) SQ is perpendicular to SP. :confused:
c) The line OP and the line through S perpendicular to the tangent at P intersect on the directrix. :confused:


Thanks ahead for any help.


clues are welcome too.
 

haboozin

Do you uhh.. Yahoo?
Joined
Aug 3, 2004
Messages
708
Gender
Male
HSC
2005
c0okies said:
The equation of the tangent of p(x1,y1) on the ellipse E:

x^2/a^2 + y^2/b^2=1 is xx1/a^2 + yy1/b^2 =1

It meets the major axis in T and a directrix in Q, PN is the ordinate of P, S is the focus corresponding to this directrix. Show that:
a) ON.OT=a^2 (done)
b) SQ is perpendicular to SP. :confused:
c) The line OP and the line through S perpendicular to the tangent at P intersect on the directrix. :confused:


Thanks ahead for any help.


clues are welcome too.

this is a standard textbook question, make sure u can do this!

b, find the coordinates of Q (i assume u know how to do that since u did a) then find its gradient
find the gradient of SP

and show that

m1.m2 = -1 THUS IT IS PERPENDICULAR.

c.
well the line which goes through the tangent and is perpendicular with it BUT IT ALSO GOES THROUGH S would be such a line

y - y1 (y coordinate of S) = m(gradient)(x - x1)

(yr 9-10 work)

now to find the gradient

u do the reverse of

m1m2 = -1

so m2 = -1/m1

and what is m1?
it is the gradient of the tangent (which u pretty much know from A)

now u found this line

and u knwo the line OP simple gradient y1/x1

so its basicly y = y1x/x1

SOLVE SIMULTANIOUSLY with the line u found earlier,

NOTE: this is very easy, u only need the X coordiantes

because if the x coordiantes are a/e then it lies on the directrix
 

Trev

stix
Joined
Jun 28, 2004
Messages
2,037
Location
Pine Palace, St. Lucia, Brisbane.
Gender
Male
HSC
2005
b)
Find the point Q simultaneously with equation of directrix and equation of tangent.
Find gradient of PS and gradient of SQ using gradient formula from the point P (given) and points S, Q (work them out).
Show m<sub>1</sub>m<sub>2</sub>=-1. ie. gradient of SP times gradient of SQ is -1.

c)
Work out equation of OP, find the y value for where this intersects the directrix. Find the gradient between this point and point S then show this is perpendicular to gradient of the tangent (implicitly differentiate equation of ellipse to find gradient, one must be the negative reciprocal of the other).
 

_ShiFTy_

Member
Joined
Aug 7, 2005
Messages
185
Gender
Male
HSC
2006
Just a quick question about Q2. I just did it, but is there a faster way apart from subbing in x=a/e into the tanget to find the y value of Q...then doing all that gradient stuff?
 

haboozin

Do you uhh.. Yahoo?
Joined
Aug 3, 2004
Messages
708
Gender
Male
HSC
2005
_ShiFTy_ said:
Just a quick question about Q2. I just did it, but is there a faster way apart from subbing in x=a/e into the tanget to find the y value of Q...then doing all that gradient stuff?

no man
unless u can point a definition to me

this could be done in relatively short steps, it could be a 2-3 mark question so it is worth it.
 

haboozin

Do you uhh.. Yahoo?
Joined
Aug 3, 2004
Messages
708
Gender
Male
HSC
2005
c0okies said:
conics rarely has shortcuts =S

actually, it is pretty much 1 of the only subjects with the MOST shortcuts.

refer to the next conics quesiton asked that i just answered.
we used a definition to get there in 1 line.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top