0

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?

Daniel Fischer
  • 206,697
roni
  • 3

2 Answers2

0

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 $

Narasimham
  • 40,495
  • I think i got your hint but not completely: the parabola is $y^2=2px$ so $2yy'=2p$ so $y'=\frac{p}{y}=\frac{1}{t^2}$ but that's not $t$ as you wrote. also if you take the opposite from your substitute: $x=pt^2, y=2pt$ so we get: $y'=\frac{1}{t}$ and that also isn't $t$ but it's simpler than the last one. but still, why the slope, as you said, is $(t_1+t_2)/2$? in the question i only know that the slope of the the line that passes in $F$ is $2$.. – roni Feb 26 '15 at 02:02
0

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

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

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

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

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

$p^2=36$

$p=6$

tomi
  • 9,594