Question

Equation of Tangent to the curve 4x²+9y²=40 at (3, 2) is​

Answer 1

Differentiate implicitly

8x + 18y (dy/dx) = 0

dy/dx = (-8x)/(18y) = -4x/(9y) = -2/9

-4x/y = -2

x/y = 1/2

x = (1/2) y

Substitute into original equation

4(1/2 y)^2 + 9y^2 = 40

y^2 + 9y^2 = 40

10y^2 = 40

y = +/- 2

So x = (1/2)(+/- 2) = +/- 1

At (1, 2), the tangent's equation is

y - 2 = (-2/9)(x - 1)

y - 2 = (-2/9)x + 2/9

y = (-2/9)x + 20/9

At (-1, -2), the tangent's equation is

y + 2 = (-2/9)(x + 1)

y + 2 = (-2/9)x - 2/9

y = (-2/9)x - 20/9