Locus in conics (1 Viewer)

[sonic1988]

Set up a coordinate axes and a line, ab, 6 cm long. point P divides ab in the ration 1:2

a/ show that the locus of P is an ellipse and write its equation
b/ what would be the locus of p if p were the midpoint of ab

 

[sonic1988]

of course i know that method, but i have no clue for that question, so thats why i am asking for helping

 

[sonic1988]

i know that e of circle is 1. But with a, i couldnt find out the equation although i knew it would turn to be an ellipse. Please help

 

[sonic1988]

ty

 

[sonic1988]

This is my answer for a. click on that link. and y^2 istead of y. Using pythagoras theorem

 

[kony]

the eccentricity of a circle is 0, not 1.

(eccentricity of 1 = parabola)

 

[elseany]

unless im missing something this question is pretty stupid... ><

were given a line ab, but it has no bounaries, this line can exist anywhere and then theres a point on this line which cuts it in the ratio 1:2. We're told to find the locus of this line that has no boundaries. The question is stupid, the locus is a null function (or a full function? :S)

What i suspect the question means is let a and b be points on the co-ordinate axis, which in that case would be a pretty standard question;



After using similair triangles to find values for the sides, just sub into pythagoras:

y^2 + (xb/a)^2 = b^2

rearrange to give;

x^2/a^2 + y^2/b^2 = 1

did anyone else think of this after looking at that question or am i way off? (im most probably wayyy off!)

 

[jyu]

Set up a coordinate axes and a line, ab, 6 cm long. point P divides ab in the ration 1:2

a/ show that the locus of P is an ellipse and write its equation
b/ what would be the locus of p if p were the midpoint of ab

:wave:

 

[sonic1988]

i dont know, i got a circle formula but when i test for e, it came out e=1 instead of e=0, thus it has to be a parabola. I cant explain about this case, plz some one figure out this problem plz

 

[elseany]

its actually a circle...

if you can be bothered theres a way to test it, if you still have your whiteboard from year 12 (if you had one...) then whip it out and sticky tape a pen half way down a ruler and then move the ruler along the edges of your whiteboard.

 

[sonic1988]

yeh, i got the same answer as 3unitz

 

[elseany]

lmao <3 3unitz ^_^

 

[haque]

For these q sonic-use polar coordinates-since p divides it in that ratio, we have AP=2 and PB=4 if @ is the acute angle this line makes with the x axis then we represent the x cordinate of p as x=2cos@ and y=4sin@ using sin^2@ +cos^2@=1 u get x squared on 4 plus y squared on 16=1. If it was the mdipoint of the line then the locus would be a CIRCLE, with x=3cos@ and y=3cos@ in the polar coordinate form. 3unitz is correct.