Let Y = Y ( X ) be a curve lying in the first quadrant such that...

Plotting the line $Y - y = {Y}^{'} ( x ) ( X - x )$ we get,

 

Now, from the above diagram,

The area of the triangle formed will be,

$A = \dfrac{1}{2}\left(\dfrac{- y}{{Y}^{'} ( x )} + x\right) ( y - x{Y}^{'}\left(x\right) )$

Now, according to the given condition we get,

$\dfrac{1}{2}\left(\dfrac{- y}{{Y}^{'} ( x )} + x\right) ( y - x{Y}^{'}\left(x\right) ) = \dfrac{- {y}^{2}}{2{Y}^{'} ( x )} + 1$

$⇒ \left(- y + x{Y}^{'} ( x )\right)\left(y - x{Y}^{'} ( x )\right) = - {y}^{2} + 2{Y}^{'} ( x )$

$⇒ - {y}^{2} + xy{Y}^{'} ( x ) + xy{Y}^{'} ( x ) - {x}^{2}{\left[{Y}^{'} ( x )\right]}^{2} = - {y}^{2} + 2{Y}^{'} ( x )$

$⇒ 2xy - {x}^{2}{Y}^{'} ( x ) = 2$

$⇒ \dfrac{dy}{dx} = \dfrac{2xy - 2}{{x}^{2}}$

$⇒ \dfrac{dy}{dx} - \dfrac{2}{x}y = \dfrac{- 2}{{x}^{2}}$

It is a linear differential equation,

Now, finding $IF = {e}^{- 2\ln x} = \dfrac{1}{{x}^{2}}$

Now, the solution is given by,

$y \cdot \dfrac{1}{{x}^{2}} = \dfrac{2}{3}{x}^{- 3} + c$

Given at $x = 1, y = 1$

$⇒ 1 = \dfrac{2}{3} + c ⇒ c = \dfrac{1}{3}$

Hence, the equation of the curve will be,

$Y = \dfrac{2}{3} \cdot \dfrac{1}{X} + \dfrac{1}{3}{X}^{2}$

$⇒ 12Y ( 2 ) = \dfrac{5}{3} \times 12 = 20$