Question

Let the area enclosed by the $x$-axis, and the tangent and normal drawn to the 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 ______

Answer 1

The correct answer is 170
4x3−3xy2+6x2−5xy−8y2+9x+14=0 at P(−2,3)
12x2−3(y2+2yxy′)+12x−5(xy′+y)−16yy′+9=0
48−3(9−12y′)−24−5(−2y′+3)−48y′+9=0
y′=−9/2
Tangent y−3=−29​(x+2)⇒9x+2y=−12
Normal :y−3=92​(x+2)⇒9y−2x=31

Area =21​(231​−4)×3=485​
8A=170