showing that 3 normals can be drawn to parabola

**differentiate** · January 15th 2010, 07:51 PM

A parabola has the parametric equations x = 2at and y = at^2. The normal to the parabola when t= p is $x + py = 2ap + ap^3.$ Show that through $(x_1, y_1)$ it is possible to draw up to three normals to the parabola

Thanks

**Isomorphism** · January 15th 2010, 08:56 PM

Originally Posted by differentiate

A parabola has the parametric equations x = 2at and y = at^2. The normal to the parabola when t= p is $x + py = 2ap + ap^3.$ Show that through $(x_1, y_1)$ it is possible to draw up to three normals to the parabola

Thanks

Geometry: The point $(x_1, y_1)$ lies on the line $x + py = 2ap + ap^3.$ for some $p$ .
Algebra: Substitute $(x_1, y_1)$ in $x + py = 2ap + ap^3.$

Geometry: To prove that it is possible to draw up to three normals to the parabola $x^2 = 4ay$ of the form $x + py = 2ap + ap^3.$ .
Algebra: To prove that it is possible that there are up to three solutions $p_1,p_2,p_3$ for the equation $x_1 + py_1 = 2ap + ap^3.$ .

Now that we have formulated the problem algebraically, can you tell why "there are up to three solutions $p_1,p_2,p_3$ for the equation $x_1 + py_1 = 2ap + ap^3.$ "?

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