Mat 3.5: Implicit Differentiation

Question

Find an equation of the tangent line to the curve

2(x^2+y^2)^2 = 25(x^2-y^2)

(a lemniscate) at the point (3,1)

An equation of the tangent line to the lemniscate at the given point is (blank).

Answer 1

2(x²+y²)² = 25(x²-y²)

below I've used y’ to represent dy/dx

differentiation of both sides gives

4 (x² + y²)(2x + 2yy’) = 25(2x – 2yy’)

8 (x³ + x²yy’ + xy² + y³y’) = 50(x – yy’)

8 x²yy’ + 8y³y’ + 50yy’ = 50x – 8(x³ + xy²)

Solving for y’

y’= [50x – 8(x³ + xy²)]/[8(x²y + y³) + 50y]

= [25x – 4(x³ + xy²)]/[4(x²y + y³) + 25y]

Plug in x =3, y =1 to get the slope (m) of the tangent line

m = [25(3) – 4((3)³ + (3)(1))]/[4((3)²(1) + (1)) + 25(1)] =

[75 – 4(27+ 3)]/[4((9 + 1) + 25] = (75 – 120)/(40 +25) = –45/65 = –9/13

So the equation for the tangent line at (3,1) is

y – 1 = (–9/13)(x – 3) in point-slope form

or

y = (–9/13)x +40/13 in slope-intercept form

Desmos graph: desmos.com/calculator/n1pqkcozyq

Hope this helps!

Answer 2

Plug in x = 3 and y = 1 to get the value of dy/dx which is the slope of the line.

Plug in the value of the slope as "m" and x₁ = 3 and y₁ = 1 into the point-slope form of a line:

y - y₁ = m(x - x₁)