Question

Let the area enclosed by the \( x \)-axis, and the tangent and normal drawn to the WV. curve \( 4 x^{3}-3 x y^{2}+6 x^{2}-5 x y-8 y^{2}+9 x+14=0 \) at the point \( (-2,3) \) be \( A \). Then \( 8 A \) is equal to [JEE Mains-2022] Ans.:(170)

Answer 1

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),

⇒ 48 – 27 + 36y' – 24 – 15 + 10y' – 48y' + 9 = 0

⇒ 2y' = –9

So, m₁ = \(\frac{-9}2\) and m₂ = \(\frac 29\)

The equation of tangent and normal is shown below:

T ≡ y – 3 = \(\frac{-9}2 (x + 2)\)

Put, y = 0

x = \(\frac{-4}3\)

N ≡ y – 3 = \(\frac 29 (x + 2)\)

Put, y = 0

x = \(\frac{-31}2\)

Therefore, area = \(\frac12\) × Base × Height

A = \(\frac 12 \times \left(\frac{-4}3 + \frac{31}2\right)(3)\)

\(= \frac 12 (\frac {85}6)(3)\)

\(= \frac{85}4\)

8A = 170