Question: Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0.
The tangent plane to the surface z=f(x,y)=xy^3+8/y is given by
2x+7y+2z=D
Where D is some constant.
The equation of the tangent plane to the surface f(x,y), at the point (x_0 y_0 ) can be written as
z=f(x_0,y_0 ) + f_x (x_0 y_0 )∙(x-x_0 ) + f_(y ) (x_0,y_0 )∙(y-y_0 )
Where (x_0,y_0) is some point on that plane andf_x (x_0 y_0 ) is the slope of the tangent line to the surface at that point in the x-direction and similarly for f_(y ) (x_0,y_0 ).
We have f(x,y)=xy^3+8/y.
f_x=y^3, f_y=3xy^2-8/y^2 .
So,
f_x (x_0 y_0 )=y_0^3, f_(y ) (x_0,y_0 )=3x_0 y_0^2-8/(y_0^2 ), f(x_0,y_0 )=x_0 y_0^3+8/y_0 .
Substituting for the above into the equation of the tangent plane,
z = x_0 y_0^3+8/y_0 + y_0^3∙(x-x_0 ) + (3x_0 y_0^2 - 8/(y_0^2 ))∙(y-y_0 )
z = x_0 y_0^3+8/y_0 + y_0^3 x - x_0 y_0^3 + (3x_0 y_0^2 - 8/(y_0^2 ))y - 3x_0 y_0^3 + 8/y_0
z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {x_0 y_0^3 + 8/y_0 - x_0 y_0^3 - 3x_0 y_0^3 + 8/y_0 }
z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {16/y_0 - 3x_0 y_0^3 }
-2y_0^3 x - 2(3x_0 y_0^2 - 8/(y_0^2 ))y + 2z = 32/y_0 - 6x_0 y_0^3
We have just written down the equation of the tangent plane, which has also been written down as
2x+7y+2z=D
Comparing the two forms of these equations,
-2y_0^3 = 2 ----------------------------------- (1)
2(3x_0 y_0^2 - 8/(y_0^2 )) = 7 ---------- (2)
32/y_0 - 6x_0 y_0^3 = D ------------------- (3)
From (1), y_0 = -1
Substituting for y_0=-1 into (2), we get x_0 = 3/2.
Substituting for x_0 and y_0 into (3), we get D = -23, and z = f(x_0,y_0 ) = (3/2) (-1)^3 + 8/(-1) = (-9.5)
The equation of the tangent plane is thus
2x+7y+2z=-23
And this plane is tangential to the surface f(x,y) at one point only, which is: (1.5,-1,-9.5)