need help starting this problem.

[stars584]

The figure shows a point P on the parabola y=x^2 and the point Q where the perpendicular bisector of OP intersects the y-axis. As P approaches the orgin along the parabola, what happens to Q? Does it have a limiting position? If so, find it.

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[pka]

The equation of the prependicular bisector is

\(\displaystyle \L y = \frac{{ - x}}{p} + \frac{1}{2} + \frac{{p^2 }}{2}.\)

For any p, q is the y-intercept.

 

[soroban]

Hello, stars584!

This is a classic problem . . . with a surprising punchline.

The figure shows a point \(\displaystyle P\) on the parabola \(\displaystyle y\,=\,x^2\)
and the point \(\displaystyle Q\) where the perpendicular bisector of \(\displaystyle OP\) intersects the y-axis.
As \(\displaystyle P\) approaches the orgin along the parabola, what happens to Q?
Does it have a limiting position? If so, find it.

Let point \(\displaystyle P\) have coordinates \(\displaystyle (p,\,p^2)\)

The slope of \(\displaystyle OP\) is: \(\displaystyle \:m\:=\:\frac{p^2\,-\,0}{p\,-\,0} \:=\\)

The midpoint of \(\displaystyle OP\) is: \(\displaystyle \,M\left(\frac{1}{2}p,\,\frac{1}{2}p^2\right)\)

The perpendicular bisector of \(\displaystyle OP\) has point \(\displaystyle M\) and slope -\(\displaystyle \frac{1}{p}\)

Its equation is: \(\displaystyle \:y\,-\,\frac{1}{2}p^2\:=\:-\frac{1}{p}\left(x\,-\,\frac{1}{2}p\right)\;\;\Rightarrow\;\;y\:=\:-\frac{1}{p}x\,+\,\frac{1}{2}\left(1\,+\,p^2\right)\)

The \(\displaystyle y\)-intercept of \(\displaystyle MP\) is: \(\displaystyle \:Q \:=\:\frac{1}{2}\left(1\,+\,p^2\right)\)


Therefore: \(\displaystyle \L\:\lim_{p\to0}\,\frac{1}{2}\left(1\,+\,p^2\right) \;=\;\frac{1}{2}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Surprised? . . . This is contrary to "common sense".

As the chord \(\displaystyle OP\) becomes "more horizontal"
. . its perpendicular bisector becomes "more vertical".

We would expect its \(\displaystyle y\)-intercept to slide up the \(\displaystyle y\)-axis ... forever.

But it doesn't . . . It stops at \(\displaystyle y\,=\,\frac{1}{2}\)
. . (This is twice the focal distance of the parabola.)

Why? . . . I don't have a clue . . .

 

[stapel]

This is a classic problem . . . with a surprising punchline....

[complete worked solutin]

Therefore: [answer]

I wonder if the student might have learned more by working on some portion of this exercise...? I wonder if perhaps the student asked for help with starting this exercise because he/she wanted to learn how to do this on his/her own...?

Too bad the student didn't get that chance....

Eliz.