Finding the value of $p$ in the parabola $y^2=2px$

Question

I just started to learn the parabola shape and I have a question:

Given the parabola $y^2=2px$ $(p>0)$.

The chord $AB$ of the parabola passes through the focus $F(\frac{p}{2},0)$.

The slope $m$ of chord $AB$ is $m_{AB}=2$.

The length of $AB$ is $|AB|=15$.

I need to find the value of $p$ (the equation of the parabola).

My Attempt

let $A(x_1,y_1),B(x_2,y_2)$ so $m_{AB}=m_{AF}=m_{BF}=2$.

Also $|AB|$ is equal to the sum of the radius $r_1+r_2$, so $x_1+x_2+p=15$.

From the slopes I got: $y_1=2x_1-p,y_2=2x_2-p$.

Then I got stuck in no going anywhere algebra.

Is my attempt ok? Or could it be done in a better way? And how can I proceed?

Answer 1

For the chord you have a gradient 2 and a fixed point $(\frac p 2,0)$

$y=2x+c$

$0=p+c$

$c=-p$

Chord $y=2x-p$ meets curve $y^2=2px$ when $(2x-p)^2=2px$

$4x^2-4px+p^2=2px$

$4x^2-6px+p^2=0$

It's not nice:

$x_1=\frac {3+ \sqrt 5} 4 p$ and $x_2=\frac {3- \sqrt 5} 4 p$

$y_1=\frac {3+ \sqrt 5} 2 p -p = \frac {1+ \sqrt 5} 2 p$ and $y_2=\frac {3- \sqrt 5} 2 p -p = \frac {1- \sqrt 5} 2 p$

$(x_1 - x_2)^2+(y_1-y_2)^2=225$

$(\frac {2 \sqrt 5} 4 p)^2+(\frac {2 \sqrt 5} 2 p)^2=225$

$\frac 5 4 p^2+5 p^2=225$

$\frac {25} 4 p^2=225$

$p^2=36$

$p=6$

Answer 2

HINT:

Parametric form is much easier.

$$ x= 2 p t , y = p t^2 $$

Slope at any point = t, slope connecting two points with slopes $t_1$ and $t_2$ is given $(t_1 +t_2)/2$.

Find distance between them in terms of $t_1$ $t_2$ and p and solve for $ t_1,t_2,p $