Consider the curve C in the xy-plane given by the equation?

Question

x^2y^2-6xy=27
find a point (a,b) on the C with a>0 at which the tangent to C has the slope m=-1. give the coordinates of the point below:
a=________
b=________

Answer 1

First find the implicit derivative dy/dx

d[x^2y^2]/dx - d[6xy]/dx = d[27]/dx

2xy^2 + 2yx^2dy/dx - 6xdy/dx - 6y = 0

dy/dx(2yx^2 - 6x) = 6y - 2xy^2

dy/dx = (y/x)(3 - xy)/(xy - 3)

dy/dx = -y/x

Since we want the tangent to this curve to be -1 then we set dy/dx=-1

and so

-1 = -y/x --> x=y
So we want any point of the form (a,a) which satisfies the equation.

a^4 - 6a^2 - 27 = 0

(a^2 + 3)(a^2 - 9) = 0

so a^2 = -3 or a^2 = 9

a^2 = -3 has no real solutions

a^2 = -9 is solved by a = 3 or a = -3

But the condition a>0 is violated for a = -3 so the only solution is a = 3. And since x=y we have a=b and so b=3.

The solution is then

a=3
b=3

Answer 2

2xy^2+2x^2*y*y´-6y -6xy´=0 so y´= (6y-2xy^2)/(2x^2y-6x)=-1
6y-2xy^2 +2x^2y-6x=0
6(y-x)+2xy( x-y)=0 so
(x-y)*( 2xy-6)=0
x=y
so x^4-6x^2-27 =0 x^2=(( 6+-sqrt(36 +108))/2
Only the + sign can be used as x^2>=0
so x^2= 9 and x=3 as a must be >0 y=3
Let´s check xy-3 =0 so y=3/x
9-18 is not 27 so there is no solution from this condition

Answer 3

(ab)^2-6(ab) - 27 = (ab+3)(ab-9) = 0
ab = -3, 9
Differentiate x^2y^2-6xy=27 at (a,b), at which y' = -1, and set the result equal to 0,
2ab^2+2a^2b(-1) - 6(b-a) = 2(b-a)(ab-3) = 0
a = b = sqrt(9) = 3, since a > 0, ab ≠ 3
Therefore the only solution is (a,b) = (3, 3).

Answer 4

taking deriv.of both sides
2x(y^2)+2(x^2)yy'-6y-6xyy'=0
y'=[6y-2xy^2]/[2x^2y-6xy]
y'=[3-xy]/[x^2-3x],
for x=a,y=b we have y'=-1
[3-ab]/[a^2-3a]=-1
finally a^2-ab-3a+3=0 (1)
Also (a,b) belongs to the curve
a^2b^2-6ab=27
Putting ab=y
y^2-6y-27=0
y=9 or y=-3
For ab=9 (1) becomes
a^2-3a-6=0
a=[3+sqrt(33)]/2 (a>0)
and b=(3/4)[sqrt(33)-3]
For ab=-3
a^2-3a+6=0
has no roots