Question

Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve 4x³ – 3xy² + 6x² – 5xy – 8y² + 9x + 14 = 0 at the point (–2, 3) be A. Then 8A is equal to ______.

Options

• 150

• 160

• 170

• 180

Answer 1

Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve 4x³ – 3xy² + 6x² – 5xy – 8y² + 9x + 14 = 0 at the point (–2, 3) be A. Then 8A is equal to 170.

Explanation:

Given curve is 4x³ – 3xy² + 6x² – 5xy – 8y² + 9x + 14 = 0.

Differentiate both sides,

12x² – 3y² – 6xyy' + 12x – 5y – 5xy' – 16yy' + 9 = 0

Satisfy the point (–2, 3),

`\implies` 48 – 27 + 36y' – 24 – 15 + 10y' – 48y' + 9 = 0

`\implies` 2y' = –9

So, m₁ = `(-9)/2` and m₂ = `2/9`

The equation of tangent and normal is shown below:

T ≡ y – 3 = `(-9)/2(x + 2)`

Put, y = 0

x = `(-4)/3`

N ≡ y – 3 = `(2)/9(x + 2)`

Put, y = 0

x = `(-31)/2`

Therefore, area = `1/2` × Base × Height

A = `1/2 xx ((-4)/3 + 31/2)(3)`

= `1/2(85/6)(3)`

= `85/4`

8A = 170