Hi
The equation you have chosen for the parabola is
is on the parabola.
But S and T are not on the parabola.
S is the focus so
T is on the directrix so
You have to find T abscissa to prove that (TS) is parallel to the tangent at P.
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
I can't seem to get gradient PT = gradient TS
Help please
Thanks in advance
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
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
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: , S(0, a),
The slope of the tangent in P is: Use the first derivation to calculate the slope of the tangent.
The slope of the line ST is.
Thus both lines must be parallel.