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QJEEM22Ma7ab6: JEE Main 2022 Mathematics - Class 12 JEEM22Ma7ab6 Solution: Let the area enclosed by the x-axis, and the tangent and normal …

Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve 4x3−3xy2+6x2−5xy−8y2+9x+14=04{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0 at the point ( −- 2, 3) be A. Then 8A is equal to ______________.

Step-by-Step Verified Solution

amp;4x3−3xy2+6x2−5xy−8y2+9x+14=0 at P(−2,3)amp;12x2−3(y2+2yxy′)+12x−5(xy′+y)−16yy′+9=0amp;48−3(9−12y′)−24−5(−2y′+3)−48y′+9=0amp;y′=−9/2amp; Tangent y−3=−92(x+2)⇒9x+2y=−12amp; Normal : y−3=29(x+2)⇒9y−2x=31\begin{aligned}& 4 x^3-3 x y^2+6 x^2-5 x y-8 y^2+9 x+14=0 \text { at } P(-2,3) \\\\& 12 x^2-3\left(y^2+2 y x y^{\prime}\right)+12 x-5\left(x y^{\prime}+y\right)-16 y y^{\prime} + 9=0 \\\\& 48-3\left(9-12 y^{\prime}\right)-24-5\left(-2 y^{\prime}+3\right)-48 y^{\prime}+9 =0 \\\\& y^{\prime}=-9 / 2 \\\\& \text { Tangent } y-3=-\frac{9}{2}(x+2) \Rightarrow 9 x+2 y=-12 \\\\& \text { Normal : } y-3=\frac{2}{9}(x+2) \Rightarrow 9 y-2 x=31\end{aligned}

JEE Main 2022 (Online) 25th July Evening Shift Mathematics - Area Under The Curves Question 66 English Explanation
amp; Area =12(312−4)×3=854amp;8 A=170\begin{aligned}& \text { Area }=\frac{1}{2}\left(\frac{31}{2}-4\right) \times 3=\frac{85}{4} \\\\& 8 \mathrm{~A}=170\end{aligned}

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