Parametrics Question (1 Viewer)

[apollo1]

could someone help me with interpreting the last part of the question.

 

[DVDVDVDV]

So you have two points on a parabola, P and Q, that are connected with a chord with a gradient ((p + q)/2). P and Q can be anywhere on the parabola as long as the gradient of the connection between them stays the same. Both P and Q have a normals that intersect at N. Since P and Q move, it follows that N does too. You need to find the Cartesian equation that plots the path that N would follow.

 

[DVDVDVDV]

Here's the solution

You know that the equations of each normal are

y = px - ap^2 and
y = qx - aq^2

Thus it follows that

2y = (p + q)x - a(p^2 + q^2)
2y = 2mx - 4am^2 - 2b
y = mx - 2am^2 - b which is the equation of a straight line

 

[apollo1]

Here's the solution

You know that the equations of each normal are

y = px - ap^2 and
y = qx - aq^2

Thus it follows that

2y = (p + q)x - a(p^2 + q^2)
2y = 2mx - 4am^2 - 2b
y = mx - 2am^2 - b which is the equation of a straight line

those arent the equation of normals, they are equation of tangents. also u didnt verify that the straight line locus is normal to parabola.

 

[math man]

you should have N equal to :


set x and y to from these new parametric eqns:



now as m is fixed we dont need to sub for m, what we will do is sub for 2 as follows:




now change y as follows:



and we can conclude this is a straight line now. To prove it is a normal we need to do:



so all i did above was show the gradient of the normal at the point N produces the same gradient as my straight line...therefore the straight line is a normal to the parabola