The graph of the function (x^2 + y^2)^2 = 4x^2y is a double folium as shown below.
(a) Find, algebraically, all points on the curve with y = 1?
(b) Verify that the slopes of tangent lines to both points with y = 1 is equal to 0?

Question

Paisley Johnson · Accepted Answer

We have: # (x^2+y^2)^2  = 4x^2y # Part (A) # y=1 => (x^2+1)^2  = 4x^2 # # :. x^4+2x^2+1 = 4x^2 #
# :. x^4-2x^2+1 = 0 #
# :. (x^2-1)^2 = 0 #
# :. x^2-1= 0 #
# :. x^2= 1 #
# :. x \ = +-1 # So the coordinates are #(1,1)# and #(-1,1)#. Part (B)
Differentiating implicitly using the chain rule and product rule we get: # 2(x^2+y^2)(2x+2ydy/dx) = (4x^2)(dy/dx) + (8x)(y) # We don't need to find an explicit expression for #dy/dx#, just its value when #x=+-1# and #y=1#.  Substituting #x^2=1# and #y=1# gives: # 2(1+1)(2x+2dy/dx) = (4)(dy/dx) + 8x #
# :. 4(2x+2dy/dx) = 4dy/dx + 8x #
# :. 2x+2dy/dx = dy/dx + 2x #
# :. 2dy/dx = dy/dx  #
# :. dy/dx = 0 \ \ \ \# QED

Axel Alvarez · Answer

undefined (a) The points on the curve with y = 1 can be found by substituting y = 1 into the equation (x^2 + y^2)^2 = 4x^2y. Simplifying the equation, we get (x^2 + 1)^2 = 4x^2. Expanding and rearranging, we have x^4 + 2x^2 + 1 = 4x^2. Combining like terms, we get x^4 - 2x^2 - 3 = 0. This is a quadratic equation in terms of x^2. Solving this equation, we find that x^2 = -1 or x^2 = 3. Taking the square root of both sides, we get x = ±√(-1) or x = ±√(3). Therefore, the points on the curve with y = 1 are (√(-1), 1), (-√(-1), 1), (√(3), 1), and (-√(3), 1).

(b) To verify that the slopes of tangent lines to both points with y = 1 are equal to 0, we need to find the derivative of the function (x^2 + y^2)^2 = 4x^2y with respect to x. Differentiating both sides of the equation implicitly, we get 2(x^2 + y^2)(2x) = 8xy + 4x^2(dy/dx). Simplifying, we have 4x^3 + 4x^2(dy/dx) = 8xy. Rearranging and solving for dy/dx, we get dy/dx = (4x^3 - 8xy) / (4x^2). Substituting the x-values of the points we found in part (a), we can calculate the slopes of the tangent lines. For each point, substitute the x-value into the equation dy/dx = (4x^3 - 8xy) / (4x^2) and simplify to find the slope. If the slope is equal to 0, then the verification is complete.

The graph of the function (x^2 + y^2)^2 = 4x^2y is a double folium as shown below. (a) Find, algebraically, all points on the curve with y = 1? (b) Verify that the slopes of tangent lines to both points with y = 1 is equal to 0?

The graph of the function (x^2 + y^2)^2 = 4x^2y is a double folium as shown below. (a) Find, algebraically, all points on the curve with y = 1? (b) Verify that the slopes of tangent lines to both points with y = 1 is equal to 0?