# showing that 3 normals can be drawn to parabola

• Jan 15th 2010, 08:51 PM
differentiate
showing that 3 normals can be drawn to parabola
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

(Wink)

Thanks
• Jan 15th 2010, 09:56 PM
Isomorphism
Geometry+Algebra = Analytical Geometry
Quote:

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

(Wink)

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 $x_1 + py_1 = 2ap + ap^3.$"?