ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: Nagisa on February 18, 2013, 06:36:00 pm

Title: I'm Back, lets do some maths.
Post by: Nagisa on February 18, 2013, 06:36:00 pm: Find the Locus of the midpoint $PF$ where $P$ is the point $(2ap , ap^2)$ on the parabola $x^2 = 4ay$ and the point $F$ is its focus.

Okay, lets roll
Title: Re: I'm Back, lets do some maths.
Post by: Nagisa on February 18, 2013, 10:46:41 pm: i worded it a bit bad, find the locus of the midpoint of the point PF is wat it meant
Title: Re: I'm Back, lets do some maths.
Post by: e^1 on February 19, 2013, 12:30:11 am: Correct me if I'm wrong :) (incomplete)

We first need to find the focus of the parabola. So...

$\begin{aligned} 4a(y - 0) = (x - 0)^2 \\\\ \text{The conic form of this parabola is: } \\ 4p(y - k) = (x - h)^2 \\\\ 4p = 4a \\ \therefore p = a \\\\ \text{If vertex of this parabola is at (0,0), then:} \\ \therefore \text{Focus is at } (0, a) \end{aligned}$

Now, we can use this to find the locus of the midpoint:

$\begin{aligned} \text{Midpoint of a straight line between 2 points is } (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) \\\\ (\frac{2ap + 0}{2}, \frac{ap^2 + a}{2}) = (ap, \frac{a(p^2 + 1)}{2}) \\\\ \therefore (ap, \frac{a(p^2 + 1)}{2}) \text{ is the the locus of midpoint } PF \\ \end{aligned}$
Title: Re: I'm Back, lets do some maths.
Post by: Nagisa on February 19, 2013, 12:37:34 am: very nice man, but you havent finished the question yet. the locus is an equation, in fact, the equation of the curve that the point PF moves along. this is where ive been able to get. i dont know wat to do next but seeings you thought of the same it makes me feel better.
Title: Re: I'm Back, lets do some maths.
Post by: Nagisa on February 19, 2013, 12:41:02 am: from there you can say, $x = ap$ and $y = \frac{a(p^2 + 1)}{2}$ and then sub it into somewhere
Title: Re: I'm Back, lets do some maths.
Post by: Nagisa on February 19, 2013, 09:37:14 am: i got it this morning lols. since we want it in the form of an equation and since we have the point of locus or w.e.

so we solve them simultaneously. $p = \frac{x}{a}$ so that $y = \frac{a({(\frac{x}{a}})^2 + 1)}{2}$

$2y = a(\frac{x^2}{a^2} + 1)$

$2y = \frac{x^2}{a} + a$

$2ay = x^2 + a^2$

$x^2 = 2ay - a^2$
Title: Re: I'm Back, lets do some maths.
Post by: atar9995 on April 09, 2013, 10:25:17 pm: Use integration by parts and slicing then you use de moivres theorem to integrated the area below the locus and the focal point. After you substitute and integrate sin^2 cos dx to find the integral derivative of the theorem in fermats last theorem