Question:medium
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 ______
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 ______
Updated On: Mar 31, 2026
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Correct Answer: 170
Solution and Explanation
To find the area \(A\), we first find the equations of the tangent and normal to the curve at the given point \((-2,3)\). The given equation of the curve is \(4x^3-3xy^2+6x^2-5xy-8y^2+9x+14=0\).
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
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