Parametric Parabola (1 Viewer)

[YBK]

Hey, can anyone please help me out with this question:

The normal to the parabola x^2=8y at the point P(4t, 2t^2) has equation:
x + ty = 4t + 2t^3


i) If this normal passes through the point N(12 , 18), show that t^3 - 7t - 6 = 0


ii) Deduce that 3 normals can be drawn to the parabola from the point N(12 , 18), and find the coordinates of the points where each of these normals meets the parabola.



I can do part i):
x + ty = 4t + 2t^3
12 + 18t = 4t + 2t^3
.: t^3 - 7t - 6 = 0



But I have no idea about the second part...

thanks !!

 

[icycloud]

Hey. Just solve t^3-7t-6=0, and you get t = -1, -2, 3 then sub that in P(4t, 2t^2) to get the three points.

 

[YBK]

Hey. Just solve t^3-7t-6=0, and you get t = -1, -2, 3 then sub that in P(4t, 2t^2) to get the three points.

OMG, it's t^3...


Thanks!!!

for some reason I kept seeing it as t^2 and thinking there's only 2 solutions...

 

[Riviet]

for some reason I kept seeing it as t^2 and thinking there's only 2 solutions...

Lol i did a very similar question involving deducing 3 normals.