parametrics help (1 Viewer)

[darkphoenix]

Show that the locus of M is a parabola and find its vertex and focal length, where M is (ap, a+ap^2)
How do you the focal length and vertex after I worked out to be this: x^2=ay-a^2? Can anyone help me out... thx

 

[Carrotsticks]

Factorise a so you have:



The vertex is (0,a)

Equate the coefficient of y with the standard form to get focal length.

 

[Timske]

(x-h)^2 = 4a(y-k)^2 , where ( h, k ) is the vertex.
:. x^2 = a(y - a) , ( 0 , a) is the vertex.

 

[darkphoenix]

thanks guys but still struggling with the focal length... i don't even get wat is the focal point in here...

 

[deswa1]

For M, is the y coordinate <img src="http://latex.codecogs.com/gif.latex?a+p^2\textup{ OR }(a+p)^2" title="a+p^2\textup{ OR }(a+p)^2" />

 

[darkphoenix]

The first one, a+ p^2, not (a+p)^2

 

[deswa1]

<img src="http://latex.codecogs.com/gif.latex?x=ap \\ p=\frac{x}{a}\\y=a+p^2\\ y=a+\frac{x^2}{a^2}\\ x^2=a^2y-a^3\\ x^2=a^2(y-a)\\ x^2=4(\frac{a^2}{4})(y-a)" title="x=ap \\ p=\frac{x}{a}\\y=a+p^2\\ y=a+\frac{x^2}{a^2}\\ x^2=a^2y-a^3\\ x^2=a^2(y-a)\\ x^2=4(\frac{a^2}{4})(y-a)" />

Which is a parabola with vertex (0,a) and focal length a^2/4

 

[darkphoenix]

Which is a parabola with vertex (0,a) and focal length a^2/4

Sorry, the y is a+a(p^2) But yea,I got it, thank you deswa and others!