大贈送_Secondary Easy Easy Maths!

Differentiate y^2=6x with respect to x:
2y(dy/dx)=6
dy/dx=3/y

Let P=(z^2/6,z), which is on the parabola
For (z^2/6,z), dy/dx=3/z...(*)

Slope between (1,5/2) and (z^2/6,z)=(z-5/2)/(z^2/6-1)...(**)

For P to be a tangent point, (*) is consistent with (**):
3/z=(z-5/2)/(z^2/6-1)
3(z^2/6-1)=z(z-5/2)
z^2/2-3=z^2-5z/2
z^2/2-5z/2+3=0
z^2-5z+6=0
(z-2)(z-3)=0
z=2 or z=3

Hence, the equations of tangent:
y-5/2=(3/2)(x-1) or y-5/2=(1)(x-1)
2y-5=3x-3 or 2y-5=2x-2
2y=3x+2 or 2y=2x+3

2011-03-06 13:35:51 補充：
Idea:
#1 Find dy/dx of the parabola y^2=6x
#2 Find the slope between point P (on the parabola) and (1,5/2)
#3 When the line linking P and (1,5/2) is the tangent of parabola, dy/dx at P=slope between P and (1,5/2)
#4 You can find dy/dx at P and the coordinate of P
#5 Form the equation

2011-03-06 20:41:12 補充：
這是數學問題還是附加數學問題?

2011-03-11 21:28:40 補充：
Alternatively:
Let y-5/2=m(x-1) => y=mx+(5/2-m) be the tangent
sub y=mx+(5/2-m) into y^2=6x
m^2x^2+2m(5/2-m)x+(5/2-m)^2=6x
m^2x^2+[2m(5/2-m)-6]x+(5/2-m)^2=0...(*)

For tangency, Discriminant of (*)=0
Then you can find the m and thus solve the tangent

Are there any other methods to solve this??

For me, it's quite difficult......to do this with differentiation.

It's not too easy!