# Another parabola property

• Dec 28th 2010, 06:55 AM
Soroban
Another parabola property

Point $\displaystyle \,P$ is on the parabola $\displaystyle y\,=\,x^2.$

The origin is: .$\displaystyle O(0,0).$

The perpendicular bisector of chord $\displaystyle \,OP$ has $\displaystyle \,y$-intercept $\displaystyle \,b.$

$\displaystyle \displaystyle \text{Find: }\,\lim_{P\to O} b$

Code:

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• Dec 28th 2010, 08:00 AM
emakarov
Let P have coordinates $\displaystyle (x_0,x_0^2)$. Then the center of OP has coordinates $\displaystyle (x_0/2,x_0^2/2)$. The slope of the chord is $\displaystyle x_0$, so the slope of the perpendicular is $\displaystyle -1/x_0$. Thus, the equation of the perpendicular bisector is $\displaystyle y=-x/x_0+b$, and $\displaystyle y(x_0/2)=x_0^2/2$. From here, $\displaystyle b=(x_0^2+1)/2$. Therefore, $\displaystyle \lim_{P\to O}b=1/2$ and not $\displaystyle \infty$ as I thought initially.

I found the following theorem in this PDF document.
Quote:

The perpendicular bisector of a parabola chord and the perpendicular to the axis through the center of the chord cut off a segment on the axis of length equal to the distance of the focus to the directrix.
The axis is the vertical one. Indeed, the length of the segment in question is $\displaystyle b-x_0^2/2=1/2$, the focus of the parabola $\displaystyle y=x^2$ is $\displaystyle (0,1/4)$ and the directrix is $\displaystyle y=-1/4$.