No idea on where to get good notes and all that, try your textbook or BoS or something. Its a topic that essentially looks a lot more difficult than it really is. You just need to play round with algebra a shitload but really you are using very basic principles of co-ordinate geometry...
hkgirl said:
one example question i have trouble with is:
one extremity of a focal chord in the parabola x^2=8y is t=3
find:
a) the gradient of the chord
b) the equation of the chord
c) the cartesian coordinates of the other extremity of the chord
d) te parameter t at this point
thanks thanks. hope to see reply asap??
desperate for help
one extremity of a focal chord in the parabola x^2=8y is t=3
Ok since x^2=8y is in the form x^2=4ay, (where a is 2) any point x will be at x=2at and any point y will be at y=a(t^2)
To find the co-ordinates of this extremity, its simple substitution here.
x=2at = 2x2x3 = 12
y=a(t^2) = 2x3^2 = 2x9 = 18
So extremity of focal chord is at (12, 18)
Obviously the focal chord has to go through the focus. But where should you calculate the focus to be? From you should know from your two unit work that in any parabola where x^2=4ay the distance to the focus is "a"....so the focus is at (0,2)
a) the gradient of the chord
Now youve got your two points to find the gradient its rise over run....
ie (18-2)/(12-0) = 4/3
b) the equation of the chord
You have your gradient you have a point so you can use point gradient formula.
y - 2 = 4/3 (x)
3y-6 = 4x
4x-3y+6 = 0 is your line equation of the focal chord
c) the cartesian coordinates of the other extremity of the chord
Cartesian co-ordinates just means the numbers, like (12, 18) on the other side of the focus.
You have the equation of the focal chord, you have the equation of the parabola so you just have to find the points that the chord cuts the parabola...how? by simultaneous equations...
To do this all you need is simultaneous equations that is solve
x^2=8y and
4x-3y+6 = 0 simultaneously.
Make these into
3x^2 = 24y (1)
32x - 24y + 48 = 0
24y - 32x - 48 = 0
ie
3x^2 - 32x - 48 = 0 (2)
3x^2 - 36x + 4x - 48 = 0
3x(x-12) + 4(x-12) = 0
(3x+4)(x-12) = 0
x = 12 or -4/3
put x = -4/3 into x^2=8y since you already have the y co-ordinate for x = 12
16/9 = 8y
y = 2/9
Point:
(-4/3, 2/9)
d) the parameter t at this point
y = at^2 = 2t^2
at this point...
2/9 = 2t^2
t^2 = 1/9
t = +/- 1/3
Correct me if I'm wrong
In any case even if the answer is not quite right (it took me a while and im a little rusty), it should give you a better idea of how to tackle the question..In any case, all the best, hope it helped and good luck!
