Conics Please (1 Viewer)

[conics2008]

Pluvia: Conics stop deleting this thread, even if you don't benefit other people might

This is the question.

The chord y=mx+c meets the ellipse x^2/16 + y^2 =1 at P and Q.

i) Find the midpoint M of PQ.

I started off by finding those two points by subbin in y=mx+c into y^2 and then making it look like quadratic and then finding the x values etc etc.. but answer is way different.

I got an equation (1+16m^2)x^2+(32mc)x+16(c^2-1)=0



x= -16mc+4root 16m^2+1-c^2/16m^2+1

x= -16-4root 16m^2+1-c^2/16m^2 +1

these are my x coord of P and Q ....

finding y we must sub x into y=mx+c

coord of P ( -16mc+4root 16m^2+1-c^2 / 16m^2+1 , 4mroot 16m^2+1-c^2 +c / 16m^2 +1 )

its a long coordinate.

now the other one Q its the same but has - sign for the root

hence midpoint is plus x1 and x2 /2

the mid point for x is -16/16m^2+1

bloody hell this is the answer i keep getting.. where am i going wrong..

and y i get c/16m^2+1

Thank you.

 

[roadrage75]

im not really sure what your doing.... but this is what i'd do:

after you have found this equation: (1+16m^2)x^2+(32mc)x+16(c^2-1)=0

finding the x- midpoints is easy. use sum of roots/2. ie (-b/a)/2. so the mid x point is this: -16mc/(1+16m^2) and then r sub it into the equation of the line: ie y = mx + c

 

[midifile]

This is the question.

The chord y=mx+c meets the ellipse x^2/16 + y^2 =1 at P and Q.

i) Find the midpoint M of PQ.

I started off by finding those two points by subbin in y=mx+c into y^2 and then making it look like quadratic and then finding the x values etc etc.. but answer is way different.

I got an equation (1+16m^2)x^2+(32mc)x+16(c^2-1)=0



x= -16mc+4root 16m^2+1-c^2/16m^2+1

x= -16-4root 16m^2+1-c^2/16m^2 +1

these are my x coord of P and Q ....

finding y we must sub x into y=mx+c

coord of P ( -16mc+4root 16m^2+1-c^2 / 16m^2+1 , 4mroot 16m^2+1-c^2 +c / 16m^2 +1 )

its a long coordinate.

now the other one Q its the same but has - sign for the root

hence midpoint is plus x1 and x2 /2

the mid point for x is -16/16m^2+1

bloody hell this is the answer i keep getting.. where am i going wrong..

and y i get c/16m^2+1

Thank you.

I think an easier way to find the midpoint would be to sub y=mx+c into the equation of the ellipse and rearrange to get a quadratic. You can then find the x-value of the midpoint by dividing the sum of roots (-b/a) by 2. Then sub this in to find the y value.

I havent actually tried this because I cant be bothered (i'm totally mathed out from two exams), but see if it works.

EDIT: Oh I just realised that roadrage already suggested this

 

[conics2008]

hey yeah i was thinking about sum of roots.. because those were my 2 x coordniate but i didn't want to take a risk, because i iddn't know it would work.. soo the answer is correct. thanks all

 

[RiSeAgAiNSt2538]

conics2008 u need to have a really good look at yourself. your name is conics and you ask a conics question. what is doing?

 

[tacogym27101990]

conics2008 u need to have a really good look at yourself. your name is conics and you ask a conics question. what is doing?

was thinkin the same thing

 

[conics2008]

I got the answer right hello.. just wanted to make sure of it..

i actuallly looked at a different answer. ( silly me )

=] its all good sorry about the false alarm.

 

[RiSeAgAiNSt2538]

the world goes back to spinning

 

[qmaz]

conics2008, wtf is wrong with you

1. you like conics
2. you cant even read the back of the book
3. you like conics

 

[Slidey]

Protip: conics2008 is a troll account

 

[Affinity]

wouldn't happen to be slidey would it? someone check ips

 

[conics2008]

LoLzzz Pulavia lol i just read that thing on top lol.. never really thought about this hahha..


I always delete my threads lol.. its just like my working out. i throw them in the bin when i finish class. I dont really have a book for maths. i just grab some scrap papers and do them there. also most of my working out are done on the board. lol..

funny guys lol..

 

[Slidey]

wouldn't happen to be slidey would it? someone check ips

And here I thought you were in touch with reality.