The problem is: at what point does the normal to at intersect the parabola a second time?
Then can I just take the slope of 8, inverse it to be and then put it in point slope form,
This ends up being .
Then do I make it ? The math gets pretty hairy after that, and I just would like to know if I'm going in the right direction!
Ah, that definitely explains my first problem.
Then the tangent line is and the normal line is or
Next step, I substitute since one of the equations is already equal to y.
Multiply across by 11 to clear fractions -> -> Divide the equation by 2 .
My next question, is the quadratic formula the next way to go? I've already got it computed, but I'm not sure what the answer means or if it's correct. I know the objective is to find the perpendicular line to the parabola and find the second intersection point. So, the quadratic formula should give me the x coordinates for the intersection points on the parabola, how do I get the f(x) coordinates?
Okay, so I get it now. I use the quad. formula and got x = 1 (makes sense, this is the point given in the question and corresponds with a output of 9), and x = -1.772727273. I plugged the latter into the original function of the parabola and got 9.252066119, so the normal line intersects the parabola a second time at (-1.772727273, 9.252066119). (1,9) is where it intersects the first time.
Is this kind of information related to finding tension in mechanics? Like if there's a swing or pendulum, could finding the intersection points of the perpendicular tell where the machine is stressed or receiving any kind of instantaneous change?