Calculus AB
posted by Annie .
Could someone please help me with these tangent line problems?
1) Find the equation of the line tangent to the given curve at the indicated point: 3y^3 + 2x^2 = 5 at a point in the first quadrant where y=1.
2) Show that there is no point on the graph of x^2  3xy + y^2 = 1 where the line tangent to the curve is horizontal.
3) Find all points on the curve xy^2  x^3y = 6 where the tangent line is vertical. Explain your reasoning.

3y^3 + 2x^2 = 5
9y^2 dy/dx + 4x = 0
dy/dx = 4x/9y^2
when y = 1 , 3 + 2x^2 = 5
2x^2 = 2
x = ± 1 , but in quadrant 1, x = 1
dy/dx = 4(1)/9 = 4/9
equation of tangent:
y1 = (4/9)(x1)
9y9 = 4x + 4
4x + 9y = 13
check:
http://www.wolframalpha.com/input/?i=plot+3y%5E3+%2B+2x%5E2+%3D+5,+4x+%2B+9y+%3D+13
notice the tangent at (1,1)
2.
x^2  3xy + y^2 = 1
2x 3x(dy/dx)  3y + 2y dy/dx = 0
dy/dx(3x + 2y) = 3y  2x
dy/dx = (3y  2x)/(2y  3x)
to have a horizontal slope, dy/dx = 0 , that is,
3y  2x = 0
y = 2x/3
sub that back into the original
x^2  3x(2x/3) + 4x^2/9 = 1
times 9
9x^2  18x^2 + 4x^2 = 1
5x^2 = 1
There is no real solution to this, so there is no such tangent.
3.
find the derivative like I showed you in the first two.
Since the tangent is vertical, the denominator of the slope must be zero.
Show that this is not possible. 
Thank you so much! And I will
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