Harder Parabola Question (1 Viewer)

[seanieg89]

One of the harder questions from my self-written quizzes for 4u tutees.

Sketch the set of all points in the cartesian plane from which three different normals can be drawn to the parabola

 

[fullonoob]

origin for the point? with just normal parabola with vertex at (0,0)

 

[seanieg89]

origin for the point? with just normal parabola with vertex at (0,0)

huh? what are you suggesting/asking?

 

[fullonoob]

only point with 3 normals is (0,0) im guessing

 

[seanieg89]

Only one normal to the parabola passes through the origin, namely the y axis.

 

[Affinity]

you're looking at something along the lines of (everything above the line)

http://www.wolframalpha.com/input/?i=plot+(3/4^(1/3))+*+(x^2)^(1/3)+++2

for a = 1. Don't think this is suitable for 4U.. essentially you need to figure out which parameters would make a cubic have 3 distinct real roots.

 

[seanieg89]

yeah thats correct.

Its a hard problem definately, but not beyond a 4u student (in theory) if properly guided.

See: http://community.boredofstudies.org/14/mathematics-extension-2/244395/polynomials.html . The converse of that result would be the ideal lead-up question.

Admittedly I did not expect any 4U student to come up with a solution immediately, but wanted to see some of their ideas so i could gauge its difficulty.