Application of Differentiation

To 香港的士司機:

The first step should be:

8x + 18y(dy/dx) = 0

as any constant function has its derivative = 0

2011-12-16 14:04:54 補充：
Let (h, k) be the point on the curve where the tangent from (3, -3) touches the curve, then:

dy/dx = -4x/(9y)

So at (h, k), dy/dx = -4h/(9k)

We have:

-4h/(9k) = (k + 3)/(h - 3)

12h - 4h2 = 9k2 + 27k

4h2 + 9k2 - 12h + 27k = 0

36 - 12h + 27k = 0

12 - 4h + 9k = 0

h = 3(3k + 4)/4

Sub this relation into:

4h2 + 9k2 = 36:

9(3k + 4)2/4 + 9k2 = 36

9(3k + 4)2 + 36k2 = 144

(3k + 4)2 + 4k2 = 16

13k2 + 24k + 16 = 16

13k2 + 24k = 0

k(13k + 24) = 0

k = 0 or -24/13

h = 3 or -30/13

N.B. Since (3, -3) is in quadrant II, the tangents are expected to touch the curve in quadrant II or III.

So for the one touching at (3, 0), the equation is x = 3

For the one touching at (-30/13, -24/13) the slope is -5/9, making up the equation:

(y + 3)/(x - 3) = -5/9

9y + 27 = 15 - 5x

5x + 9y - 12 = 0

4x^2+9y^2=36 ----------(1)

Differentiate the equation (1),
dy/dx = -4x/9y

Slope between the required (x,y) & (3,-3) = (y+3)/(x-3)

(y+3)/(x-3)=-4x/9y
After solving, x = 3(4+3y)/4

Substitute this into 1,
after solving y = 0 or y=-24/13
the corresponding values of x are 3 and -15/13 respectively (15/13 rejected)

Thus, x=3 in the first condition.
dy/dx| (-15,13, -24,13) = -5/18
y+24/13 = (-5/18)(x+15/13)
65x+234y+507=0

I think i will first let the tangents' equation be y = mx+c,
then use the point (3,-3) and discriminant to solve m and c

Differentiate 36 (a constant) with respect to x = 0

Final answer:
Equation of tangent y = (4/9) x – 13/3

By implicit differentiation
8x + 18y(dy/dx) = 36
36 - 8x = 18y(dy/dx)
dy/dx = (18 - 4x)/9y

dy/dx | x = 3, y = -3
= [18 - 4(3)]/9(-3)
= (18 - 12)/-27
= -2/9

Equation of tangent:
y + 3 = -2/9(x - 3)
y = -2/9x + 2/3 - 9/3
y = -2/9x - 7/3

Did it within a minute, i don't know if it is right or not
please tell me if there's any mistake