Parabola - Parametric to Cartesian Conversion (1 Viewer)

[gcalway]

Hi,

I am working on a problem where I know a point (-2at,at^2 +4a) lies on an inverted parabola about the y-axis. Normally I would transform x = -2at to t = -x/2a and then sub for t in y = at^2 + 4a to get to the Cartesian form. But this does not work in this instance. I think it has something to do with the x = -2at.

Can anyone explain what is happening and the correct approach to the problem?

Thanks in advance,

Graeme

 

[InteGrand]

Hi,

I am working on a problem where I know a point (-2at,at^2 +4a) lies on an inverted parabola about the y-axis. Normally I would transform x = -2at to t = -x/2a and then sub for t in y = at^2 + 4a to get to the Cartesian form. But this does not work in this instance. I think it has something to do with the x = -2at.

Can anyone explain what is happening and the correct approach to the problem?

Thanks in advance,

Graeme

 

[gcalway]

InteGrand,

Therein lies the problem. The parabola thus defined is open at the top and I know the locus of the point is on an inverted parabola as I have physically drawn a sketch to confirm the locus.

Graeme

 

[InteGrand]

InteGrand,

Therein lies the problem. The parabola thus defined is open at the top and I know the locus of the point is on an inverted parabola as I have physically drawn a sketch to confirm the locus.

Graeme

The given parameterisation will give an ''upside-down'' parabola if and only if a is negative. So maybe there's a mistake in the parameterisation or a is negative.

What was the original question?

 

[gcalway]

My question arises from the following:

The normal at any point P (2at,at^2) on the parabola x^2 = 4ay cuts the y -xis a Q and is projected to R so that PQ=QR
A) Find the coordinates of R in terms of t.

My answer at this point is R(-2at,at^2 + 4a) ) I arrived at that by identifying the equation to the normal as x + ty = at^2 + 2at. Then I subbed in x = 0 to give y = at^2 + 2a therefore Q is -2at left of and +2a above P. That means R will be -2at left of and +2a above Q.

B) Find the Cartesian equation of the locus of R

Drawing the diagram for this we have a parabola centered on the y-axis passing through (0,0) which is open at the top. Therefore if we move P to the origin then the y-axis is the normal and R will be at it's maximum. Hence my assertion that the Cartesian form must be y = -x^2....... (inverted parabola).

I think that the parametric form for a parabola assumes a > 0 and that might be a contributing factor to my dialemma. But I am unsure about that and a trying to understand the true reasoning.

Graeme

 

[leehuan]

I accidentally used p and not t.


If I use the point P(+2at, at^2) and R(-2at, at^2+4a) then by moving the slider for p I get the parabola concaves up.

 

[gcalway]

Ah, the wonders of the mighty computer over the visualisation of my mind. You are indeed correct and I have been chasing a phantom all this time. I had the correct answer all the time.

Looks like my time would have been better spent learning to use GeoGebra.

Many thanks for your graphical representation it has helped my immensely.

Graeme