differentiation(5)

(a) First sub. R and S into the curve.

12A + 2√3B + C = 16...(1)

12A - 2√3B + C = 16...(2)

(1) - (2) : 4√3B = 0 => B = 0

(b) Now, the curve is simplified as Ax^2 + Cy^2 = 16

Differentiation with respect to x on both sides

2Ax + (2Cy)(dy/dx) = 0

Ax + (Cy)(dy/dx) = 0

dy/dx = -Ax/Cy

Since the slope of the tangent to the curve at point R is -√3/2

-2√3A/C = -√3/2 => A = C

The curve is simplified as Ax^2 + Ay^2 = 16

Finally, sub. S = (-2√3, 1),

13A = 16

A = 16/13

The values of A and C are both 16/13

2011-01-14 20:40:41 補充：
Sorry, start from

-2√3A/C = -√3/2 => C = 4A

The curve is simplified as Ax^2 + 4Ay^2 = 16

Finally, sub. S = (-2√3, 1),

15A = 16

A = 1

The values of A and C are 1 and 4 respectively.

2011-01-14 20:41:12 補充：
"15A = 16" should be "16A = 1"

2011-01-14 20:41:56 補充：
"15A = 16" should be "16A = 16"

(a)
sub R and S into the eq gives:
12A + 2(sqrt3)B + C = 16
12A - 2(sqrt3)B + C = 16
suntracting the two eqs gives B=0
(b)
the eq becomes Ax^2 + Cy^2 = 16
differentiate the eq on both sides with respect to x
2Ax + 2yC(dy/dx) = 0
therefore dy/dx = -Ax/Cy
sub R intp the above eq
-2(sqrt3)A/C = -(sqrt3)/2
so 4A = C
sub R back to Ax^2 + 4Ay^2 = 16
16A=16
therefore A=1 => C=4