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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Geometry+Algebra = Analytical Geometry

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**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

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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. $"?