4 / 16 + y^2 / 25 = 1

y^2 / 25 = 3 / 4

y^2 = 75 / 4

y = +/- 5 sqrt(3) / 2

Differentiating the equation:

2x / 16 + (2y / 25)(dy/dx) = 0

dy/dx = – (x / 8)(25 / 2y)

= – 25x / 16y.

When x = 2 and y = 5sqrt(3)/2:

dy/dx = – 25 * 2 / 8 * 5 sqrt(3)

= – 5 / 4 sqrt(3)

= – 5 sqrt(3) / 12.

When x = 2 and y = – 5sqrt(3)/2:

dy/dx = 25 * 2 / 8 * 5 sqrt(3)

= 5 sqrt(3) / 12.

At (2, 5 sqrt(3) / 2), the tangent is:

y – 5 sqrt(3) / 2 = – (5 sqrt(3) / 12)(x – 2)

y = – 5x sqrt(3) / 12 + 10 sqrt(3) / 3.

At (2, – 5sqrt(3) / 2), the tangent is:

y + 5 sqrt(3) / 2 = (5 sqrt(3) / 12)(x – 2)

y = 5x sqrt(3) / 12 – 10 sqrt(3) / 3.