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Geometry help

A line y = mx bisects two distinct chords drawn from (4, 4) on y2 = 4x, find the slope of the line m

I tried hard but could not get the third condition to solve...please help to solve this problem

Reply 1

TeeEm

Original post by mnavheen

A line y = mx bisects two distinct chords drawn from (4, 4) on y2 = 4x, find the slope of the line m

I tried hard but could not get the third condition to solve...please help to solve this problem

can you post a photo/scan of the question please.

Reply 2

mnavheen

A line y = mx bisects two distinct chords drawn from (4, 4) on y^2 = 4x, find the slope of the line m. I've made and attached the figure on this problem...
Thanks

Original post by TeeEm

can you post a photo/scan of the question please.

geometry.JPG13.6KB

Reply 3

atsruser

The information doesn't seem to be enough to uniquely define m.

In your diagram, if we leave the longer chord untouched, but move the shorter chord so that it ends at (0,0), then the gradient of the bisector of the chords is clearly different.

Reply 4

TeeEm

Original post by mnavheen

A line y = mx bisects two distinct chords drawn from (4, 4) on y^2 = 4x, find the slope of the line m. I've made and attached the figure on this problem...
Thanks

very interesting problem

what I have found so far is that if the points are distinct then they must both lie on the same branch of the parabola as (4,4)

I have a bit of time this afternoon so I will look at it.

EDIT one of the points must be (X,Y) with X>4 and Y>4 and the other point with Y<4

(edited 9 years ago)

Reply 5

ghostwalker

Original post by atsruser

In your diagram, if we leave the longer chord untouched, but move the shorter chord so that it ends at (0,0), then the gradient of the bisector of the chords is clearly different.

But it won't then have the form y=mx, or did I misunderstand you?

Reply 6

Notnek

Original post by atsruser

You can't draw a line y=mx that bisects a chord passing through (0,0).

Reply 7

atsruser

Original post by notnek

You can't draw a line y=mx that bisects a chord passing through (0,0).

Right. I didn't read the question properly. I was reading a non-existent "+c".

Reply 8

Spandy

There will be a range of values for m, I believe the answer should be (-1,1). Slope of line must be less than 1 and the other boundation was obtained by the fact that if we consider the chord to intersect very near the origin, the line must be the normal.

Reply 9

ghostwalker

Couple of values are given by points on the lower arc of (16,-8) (36,-12)
Midpoints are (10,-2) and (20,-4) giving m= -1/5

and (9,-6),(100,-20) with midpoints (13/2,-1) and (52,-8) giving m= -2/13.

So, certainly, multiple solutions.

Edit: Any pairs of (x1,y1),(x2,y2) satisfying (y1+4)(y2+4)=32 gives a valid solution - maybe some restrictions, not checked.