Author Topic: I'm Back, lets do some maths. (Read 1983 times) Tweet Share

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

Nagisa · « **Reply #1 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

e^1 · « **Reply #2 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:

$\begin{aligned} \text{Midpoint of a straight line between 2 points is } (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) \\\\ (\frac{2ap + 0}{2}, \frac{ap^2 + a}{2}) = (ap, \frac{a(p^2 + 1)}{2}) \\\\ \therefore (ap, \frac{a(p^2 + 1)}{2}) \text{ is the the locus of midpoint } PF \\ \end{aligned}$

Nagisa · « **Reply #3 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.

Nagisa · « **Reply #4 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

Nagisa · « **Reply #5 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 $y = \frac{a({(\frac{x}{a}})^2 + 1)}{2}$

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

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

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

$x^2 = 2ay - a^2$

atar9995 · « **Reply #6 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

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Author Topic: I'm Back, lets do some maths. (Read 1983 times) Tweet Share

Nagisa

I'm Back, lets do some maths.

Nagisa

Re: I'm Back, lets do some maths.

e^1

Re: I'm Back, lets do some maths.

Nagisa

Re: I'm Back, lets do some maths.

Nagisa

Re: I'm Back, lets do some maths.

Nagisa

Re: I'm Back, lets do some maths.

atar9995

Re: I'm Back, lets do some maths.

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