# Thread: The Parabola in Parametric Form... Proving parallel lines!

1. ## The Parabola in Parametric Form... Proving parallel lines!

The straight line drawn from a point P on a parabola to the vertex cuts the directrix at T. Prove that if the focus is at S, that TS is parallel to the tangent at P.

S = (2as, as^2)
P = (2ap, ap^2)
T = (2at, at^2)

gradient PT = (at^2 - ap^2)/2at - 2ap
= a (t^2 - p^2)/ 2a(t-p)
= (t + p)/2

gradient TS = (at^2 - as^2)/(2at - 2as)
= (t + s)/2

2. Hi

The equation you have chosen for the parabola is $y = \frac{x^2}{4a}$
$P(2ap,ap^2)$ is on the parabola.
But S and T are not on the parabola.

S is the focus so $S(0,a)$
T is on the directrix so $T(x,-a)$

You have to find T abscissa to prove that (TS) is parallel to the tangent at P.

3. Originally Posted by running-gag
Hi

The equation you have chosen for the parabola is $y = \frac{x^2}{4a}$
$P(2ap,ap^2)$ is on the parabola.
But S and T are not on the parabola.

S is the focus so $S(0,a)$
T is on the directrix so $T(x,-a)$

You have to find T abscissa to prove that (TS) is parallel to the tangent at P.
ahhh I see.. but how would I find T abscissa?

4. Originally Posted by differentiate
ahhh I see.. but how would I find T abscissa?
Through its definition : T is the intersection point between :
- the line passing through P(2ap,apē) and the vertex of the parabola (0,0)
- the directrix of the parabola whose equation is y=-a

5. Originally Posted by running-gag
Through its definition : T is the intersection point between :
- the line passing through P(2ap,apē) and the vertex of the parabola (0,0)
- the directrix of the parabola whose equation is y=-a
so I would have to solve simultaneously the equation of point P to the vertex, and y = -a

equation Point P to the vertex
ap^2 - 0 y - 0
------- = ..........
2ap - 0 x - 0

y2ap = xap^2

y = 2ap/xap^2

y = -a

-a = 2ap/xap^2
-xa^2p^2 = 2ap
-xap = 2

now what???

soz.. I'm new to this kind of stuff

6. Originally Posted by differentiate
so I would have to solve simultaneously the equation of point P to the vertex, and y = -a

equation Point P to the vertex
ap^2 - 0 y - 0
------- = ..........
2ap - 0 x - 0

y2ap = xap^2

...
$2apy = ap^2x$ . Divide through by 2ap:

$y = \dfrac{ap^2}{2ap} x = \dfrac p2 x$

Now y = -a. Plug in this term and solve for x:

$-a=\dfrac p2 x~\implies~x = -\dfrac{2a}p$

Thus $T\left(-\dfrac{2a}p\ ,\ -a\right)$

7. Originally Posted by earboth
$2apy = ap^2x$ . Divide through by 2ap:

$y = \dfrac{ap^2}{2ap} x = \dfrac p2 x$

Now y = -a. Plug in this term and solve for x:

$-a=\dfrac p2 x~\implies~x = -\dfrac{2a}p$

Thus $T\left(-\dfrac{2a}p\ ,\ -a\right)$
ahhh thanks... okay, after that I would need to calculate the gradient of TS and tangent at P....

TS =
s + a
----------
0 + 2a/p

= ps + ap
-------------
2a

but what does it mean by tangent at P...would that mean PT? but that wouldn't make sense, because they wouldn't be parallel if it was PT and ST...

soz again. I'm so stupid

8. Originally Posted by differentiate
ahhh thanks...
...
but what does it mean by tangent at P...would that mean PT? but that wouldn't make sense, because they wouldn't be parallel if it was PT and ST...
I assume that you have made a sketch. (see attachment)

As you can see the line ST is parallel to the tangent in P at the parabola.

You already have: $P(2ap, ap^2)$ , S(0, a), $T\left(-\dfrac{2a}p\ ,\ -a\right)$

The slope of the tangent in P is: $m_t=\dfrac1{2a} \cdot 2ap= p$ Use the first derivation to calculate the slope of the tangent.

The slope of the line ST is.

$m_{ST}=\dfrac{a-(-a)}{\frac{2a}p - 0} = p$

Thus both lines must be parallel.