Sarahisme Posted September 29, 2005 Posted September 29, 2005 ahhh i don't see where i have gone wrong, i think i've made a small slip somewhere (either that or i am way off track ) anywho, heres what i did (the question is attached as a picture)... Let q = (a,b,c), so c = [math] \frac{a^2 + b^2}{4} [/math] now i get the tangent plane to P at q to be z = [math] \frac{a^2 + b^2}{4} + 0.5a(x - a) + 0.5b(y-b) [/math] and the normal at q to be n = (0.5a)i + (0.5b) -k then i call the vector qF the vector from q to the focus so i get qF to be qF = <-a,-b,1-c> then i figure all i need to do is show that (-k + qF) is parallel to the tangenet plane to P at q. That is that (-k + qF) is perpendicular to the normal vector at q. I then also need to show that (-k - qF) is orthogonal to the tangent plane. That is that (-k - qF) is parallel to the normal vector at q. To show that (-k + qF) is perpendicular to the normal vector at q i use the dot product rule so i need to show that (-k + qF).(n) = 0 likewise for (-k - qF), i need to show, using the cross product, that (-k - qF)x(n) = 0 and it almost works but i think the normal thing show have [math] \frac{1}{4} [/math]'s in it....anyway...does anyone know where i have gone wrong? or did i ever even go in the right direction to begin with? Thanks Sarah p.s. i also attached a picture of the parabaloid (in the next post) if that helps people ( i think its the right graph, i don't use maple much...) is it the right graph?
Sarahisme Posted September 29, 2005 Author Posted September 29, 2005 i hope this is this right graph....
Sarahisme Posted September 29, 2005 Author Posted September 29, 2005 nup its ok i know what i did wrong! i should have taken the unit vector qF!!!! woohoo i solved it! :):D:D:D:D:D:D
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