Question

Find the equation of the tangent and the normal to the given curve at the indicated point: 16x² + 9y² = 144 at (2, y₁), where y₁ > 0

Answer 1

Consider 16x² + 9y² = 144 as the equation of curve

Taking x₁ = 2 we get y₁ = 4√5/3

By differentiating both sides w.r.t. x

32x + 18y dy/dx = 0

We can write it as

18y dy/dx = – 32x

So we get

dy/dx = -16x/ 9y

Here

(dy/dx)_{(2, 4√5/3)} = (- 16×2)/(9 × 4√5/3) = – 8/ 3√5

Here the equation of tangent at point (2, 4√5/3)

y – 4√5/3 = – 8/ 3√5 (x – 2)

On further calculation

3√5y – 20 = -8x + 16

We get

8x + 3√5y – 36 = 0

Here the equation of the normal at point (2, 4√5/3)

y – 4√5/3 = 3√5/8 (x – 2)

It can be written as

8y – 32√5/3 = 3√5x – 6√5

On further calculation

24y – 32√5 = 9√5x – 18√5

So we get

9√5x – 24y + 14√5 = 0