Question 1
You MAY use a calculator to solve this problem.
Consider the equation x2 - 2xy + 4y2 = 64
(a) Write an expression for the slope of the curve at the point (x, y).
(b) Find the equation of the tangent lines to the curve at the point x = 2.
(c) Find 
at (0, 4).
Solution
a) differentiate implicitly with respect to x and remembering that y is a function of x
2x-[2(xy' - y(1))] +8yy'= 0
2x - 2xy' - 2y +8yy' = 0
2xy' +8yy'= 2y-2x
y'(2x-8y) = 2y-2x
dy= 2y-2x
dx 2x-8y
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b) First find the y-coordinate of the equation x2 - 2xy + 4y2 = 64 at x=2
22 - 2(2)y + 4y2 = 64
4-4y+4y2 = 64
4(y2 +2y-20) = 0
4(y+6)(y-4)=0
y=-6, y=4 (original answer: y= 6, 4)
we have two points (2,-6) and (2,4)
y' (2,-6) = 2(-6)-2(2)
2(2)-8(-6)
y' (2,6) = -4/13 <--- the slope at point (2,6) (original answer: y(2,6) = -1/6)
y'(2,4) = 2(4)-2(2)
2(2)-8(4)
y'(2,4) = -1/7 <--- slope of the line at (2,4)
I THINK I MIXED SOMETHING UP HERE.
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c) differentiate the slope equation first.
y' = 2y-2x
2x-8y
Think Lodihi-hidilo...
= (2x-8y)D(2y-2x) - (2y-2x)D(2x-8y)
(2x-8y)2
= (2x-8y)(2y'-2) - (2y-2x)(2-8y')
(2x-8y)2
= 4xy'-4x-16yy'+16y - (4y-16yy'-4x+16xy')
(2x-8y)2
= 4xy'-4x-16yy'+16y +4y+16yy'+4x-16xy'
(2x-8y)2
= 12xy'+20y
(2x-8y)2
Substitute y' into this equation.
= 12x(
2y-2x)+20y
(2x-8y)
(2x-8y)2
Simplify by finding a common denominator
= 12x(2y-2x)+20y (2x-8y)
(2x-8y) (2x+8y)
(2x-8y)2
1
= 12x(2y-2x) +20y(2x-8y)
2x+8y_______
(2x-8y)2
1
=24xy-24x2 + 40xy-160y * ( 1 )
2x-8y (2x-8y)2
= 16xy-24x2 -160y
(2x+8y)^3
= 8(2xy- 3x2 - 20y)
(2x+8y)^3
Now substitute the point (0,4) into the equation
= 8(2(0)(4)- 3(0)2 - 20(4))
(2(0)+8(4))^3
= -640
32768
= -5_
256
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