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ellipse question (1 Viewer)

c0okies

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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

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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

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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_

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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

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_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

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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.
 

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