The area (in square units) of the triangle fo | vedclass

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(B) Given the curve $x^3 + y^3 = 2xy$. Differentiating with respect to $x$,we get $3x^2 + 3y^2 \frac{dy}{dx} = 2y + 2x \frac{dy}{dx}$.At the point $(1

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(B) Given the curve $x^3 + y^3 = 2xy$. Differentiating with respect to $x$,we get $3x^2 + 3y^2 \frac{dy}{dx} = 2y + 2x \frac{dy}{dx}$.At the point $(1, 1)$,$3(1)^2 + 3(1)^2 \frac{dy}{dx} = 2(1) + 2(1) \frac{dy}{dx}$,which simplifies to $3 + 3 \frac{dy}{dx} = 2 + 2 \frac{dy}{dx}$.Thus,$\frac{dy}{dx} = -1$. The slope of the tangent $m_t = -1$ and the slope of the normal $m_n = 1$.The equation of the tangent at $(1, 1)$ is $y - 1 = -1(x - 1)$,which simplifies to $x + y = 2$.The equation of the normal at $(1, 1)$ is $y - 1 = 1(x - 1)$,which simplifies to $y = x$.The tangent intersects the $X$-axis $(y = 0)$ at $x = 2$,so the point is $(2, 0)$.The normal intersects the $X$-axis $(y = 0)$ at $x = 0$,so the point is $(0, 0)$.The triangle is formed by the vertices $(0, 0)$,$(2, 0)$,and $(1, 1)$.The base of the triangle on the $X$-axis is the distance between $(0, 0)$ and $(2, 0)$,which is $2$ units.The height of the triangle is the $y$-coordinate of the point $(1, 1)$,which is $1$ unit.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 1 = 1$ square unit.

The area (in square units) of the triangle formed by the $X$-axis,the tangent,and the normal drawn at $(1, 1)$ to the curve $x^3 + y^3 = 2xy$ is

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