Given M and N are fixed points on y axis with coordinates (0,1), (0,-1). Point P is a dynamic point on parabola y = \dfrac{1}{4}x^2

Evaluate position relationship between the line y=-1 and the circle that is centered at P with radius PM.

Let point Q is the intersecting point of the line PM with the parabola y = \dfrac{1}{4}x^2 on another side. Show that \angle PNM = \angle QNM

Collected in the board: Parabola

Steven Zheng posted 1 week ago


Let (x,y) be the coordinates of point P

Since y = \dfrac{1}{4}x^2, the coordinates could be rewritten as (x, \dfrac{1}{4}x^2). Drop altitude from point P to line y = -1. Then the coordinates of the projection point H are (x,-1)

The radius of the circle PM could be determined by using the distance formula between two points.

PM = \sqrt{x^2+( \dfrac{1}{4}x^2-1)^2} = \sqrt{( \dfrac{1}{4}x^2+1)^2} = \dfrac{1}{4}x^2+1

On the other hand, the distance between point P and the line y=-1 is simply the difference of the y coordinate of point P and H.

PH = \dfrac{1}{4}x^2 - (-1) = \dfrac{1}{4}x^2 +1



which means the line y = -1 is tangent to the circle that is centered at point P with radius PM.

Drop altitude from point Q to line y = -1 with projection point R

Since P is an arbitrary point on the parabola, the conclusion (1) is also applicable to point Q, that is


Since PH\parallel MN\parallel QR ,

\dfrac{QM}{RN} = \dfrac{MP}{NH}

Substituting (1) and (2) results in

\dfrac{QR}{RN} = \dfrac{PH}{NH}

Therefore, \triangle QRN \sim \triangle PHN

\angle HPN = \angle RQN

Noticed \angle PNM and \angle HPN are alternate interior angles, and \angle QNM and \angle HPN are also alternate interior angles.


\angle PNM = \angle QNM

Steven Zheng posted 1 week ago

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