If A is the area in the first quadrant enclosed by the curve C : 2 x 2 - y + 1 = 0 , the tangent to C at the point (1, 3) and the line x + y = 1 , then the value of 60A is ______ . [2023]

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Solution

Q.
If A is the area in the first quadrant enclosed by the curve $C : 2 x^{2} - y + 1 = 0$ , the tangent to C at the point (1, 3) and the line $x + y = 1$ , then the value of 60A is ______ . [2023]

Ans.
(16)

$Given, C : 2 x^{2} - y + 1 = 0$

$\Rightarrow 4 x - \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = 4 x \Rightarrow {[\frac{d y}{d x}]}_{(1, 3)} = 4$

$Equation of tangent to the curve C is$

$y - 3 = 4 (x - 1)$

$\Rightarrow y - 4 x + 1 = 0 ...(i)$

$Also, given line is x + y = 1 ...(ii)$

$Solving (i) and (ii), we get x = 0.4, y = 0.6$

$Given curve and equation (ii) intersect each other at (0, 1) and (- 0.5, 1.5) .$

$Required area,$

$A = \int_{0}^{1} (2 x^{2} + 1) d x - area of ∆ E O A - area of ∆ B E D + area of ∆ E D C$

$= {[\frac{2 x^{3}}{3} + x]}_{0}^{1} - \frac{1}{2} \times 1 \times 1 - \frac{1}{2} (1 - \frac{1}{4}) \times 3 + \frac{1}{2} \times (1 - \frac{1}{4}) \times 0.6$

$= (\frac{2}{3} + 1) - \frac{1}{2} - \frac{9}{8} + \frac{9}{40} = \frac{16}{60} sq. units$

$∴$ $60 A = \frac{16}{60} \times 60 = 16$

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