Implicit Differentiation #4

Question

The graph of the equation  is an ellipse lying obliquely in the plane, as illustrated in the figure below.

a. Compute .

  .

b. The ellipse has two horizontal tangents. Find an equation of the lower one.

The lower horizontal tangent line is defined by the equation   .

c. The ellipse has two vertical tangents. Find an equation of the rightmost one.

The rightmost vertical tangent line is defined by the equation   .

d. Find the point at which the rightmost vertical tangent line touches the ellipse.

The rightmost vertical tangent line touches the ellipse at the point  .

Hint: The horizontal tangent is of course characterized by . To find the vertical tangent use symmetry, or solve .

Answer 1

The graph of the equation  is an ellipse

a)

2x dx + y dx + x dy + 2y dy = 0

(2x + y) dx = -(2y + x) dy

dy/dx = -(2x + y)/2y + x)

b) The horizontal tangent is where dy/dx = 0 -(2x + y)/2y + x) = 0 when 2x + y = 0 or y = -2x

as the horizontal tangent has y as a constant, let x = -y/2

Plugging into x² + xy + y² = 5, we get (-y/2)² + (-y/2)y + y² = 5

3/4 y² = 5

y = ± 2√15/3

The lower line is y = -2√15/3

c) The vertical tangent is found by having the denominator of dy/dx = 0

2y + x = 0

From symmetry, we know the solution is x = ± 2√15/3, with the right tangent x = 2√15/3

d) y = -x/2 = -2√15/3/2 = -√15/3

The tangent point is (2√15/3, -√15/3)