Conics Midpoint Locus (1 Viewer)

[cutemouse]

Hello, I'm having a problem with a conics midpoint locus question. Thanks to anyone who can help me.

y=mx+1 is a chord of the ellipse x²+4y²=16 that passes through (0,1) for all real values of m. Find the locus of the midpoint of the chord as m varies.

I can find the midpoint in terms of m, but not sure where to go after that.

 

[azureus88]

ok i'll start from teh midpoint in terms of m part.

midpoint is given by:

x = -4m/(4m^2 + 1)
y = 1/(4m^2 + 1)

y/x = -1/4m
m = -x/4y

Sub this value of m back into equation of chord gives:

y= -x^2/4y + 1
x^2 + 4y^2 = 1 is locus

 

[cutemouse]

Hello,

Thanks for your help, but where did you get y/x from?

Also, wouldn't the chord be bounded by the ellipse, so I would need to give some sort of domain right?

 

[vds700]

Hello,

Thanks for your help, but where did you get y/x from?

Also, wouldn't the chord be bounded by the ellipse, so I would need to give some sort of domain right?

he divided the 2 equations for y and x

x = -4m/(4m^2 + 1) ......(1)
y = 1/(4m^2 + 1)...........(2)

(2)/(1) ----> y/x = -1/4m

 

[cutemouse]

Yes, I know that he divided the two equations. But I'm wondering why he did that.

Thanks for the helps btw

 

[jet]

It was just a way to eliminate the quadratic term in m. With a linear term it is much easier, plus, it allows one equation to specify the locus instead of two.
And yeah you should give a domain.

 

[cutemouse]

Thanks for your reply.

Where would this be covered? Because I don't ever recall dividing a quadratic to simplify it.

 

[jet]

Just in the worked solutions. Its a method of elimination thats is. No real hard theory.