# showing that 3 normals can be drawn to parabola

• Jan 15th 2010, 07:51 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 \$\displaystyle x + py = 2ap + ap^3. \$ Show that through \$\displaystyle (x_1, y_1) \$ it is possible to draw up to three normals to the parabola

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Thanks
• Jan 15th 2010, 08:56 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 \$\displaystyle x + py = 2ap + ap^3. \$ Show that through \$\displaystyle (x_1, y_1) \$ it is possible to draw up to three normals to the parabola

(Wink)

Thanks

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

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