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(D) Given the curve equation: $4x^{3}-3xy^{2}+6x^{2}-5xy-8y^{2}+9x+14=0$.Differentiating with respect to $x$:$12x^{2} - 3y^{2} - 6xyy' + 12x - 5y - 5x

Vedclass · Accepted Answer

$170$

Vedclass · Answer

(D) Given the curve equation: $4x^{3}-3xy^{2}+6x^{2}-5xy-8y^{2}+9x+14=0$.Differentiating with respect to $x$:$12x^{2} - 3y^{2} - 6xyy' + 12x - 5y - 5xy' - 16yy' + 9 = 0$.Substituting the point $(-2, 3)$:$12(-2)^{2} - 3(3)^{2} - 6(-2)(3)y' + 12(-2) - 5(3) - 5(-2)y' - 16(3)y' + 9 = 0$.$48 - 27 + 36y' - 24 - 15 + 10y' - 48y' + 9 = 0$.$-9 - 2y' = 0 \implies y' = -\frac{9}{2}$.Slope of tangent $m_{T} = -\frac{9}{2}$,slope of normal $m_{N} = \frac{2}{9}$.Equation of tangent: $y - 3 = -\frac{9}{2}(x + 2) \implies y = -\frac{9}{2}x - 6$. For $y=0$,$x = -\frac{4}{3}$.Equation of normal: $y - 3 = \frac{2}{9}(x + 2) \implies y = \frac{2}{9}x + \frac{31}{9}$. For $y=0$,$x = -\frac{31}{2}$.Area $A = \frac{1}{2} 	imes |x_{T} - x_{N}| 	imes |y_{P}| = \frac{1}{2} 	imes |-\frac{4}{3} - (-\frac{31}{2})| 	imes 3 = \frac{3}{2} 	imes |-\frac{8}{6} + \frac{93}{6}| = \frac{3}{2} 	imes \frac{85}{6} = \frac{85}{4}$.$8A = 8 	imes \frac{85}{4} = 170$.

Let the area enclosed by the $x$-axis,and the tangent and normal drawn to the curve $4x^{3}-3xy^{2}+6x^{2}-5xy-8y^{2}+9x+14=0$ at the point $(-2,3)$ be $A$. Then $8A$ is equal to $.......$

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