The area of the triangle formed by the normal | vedclass

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(B) Given the curve $y = e^{2x} + x^2$.First,find the derivative: $\frac{dy}{dx} = 2e^{2x} + 2x$.At the point $(0, 1)$,the slope of the tangent is $m_

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$1$

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(B) Given the curve $y = e^{2x} + x^2$.First,find the derivative: $\frac{dy}{dx} = 2e^{2x} + 2x$.At the point $(0, 1)$,the slope of the tangent is $m_t = 2e^0 + 2(0) = 2$.The slope of the normal $m_n$ is $-\frac{1}{m_t} = -\frac{1}{2}$.The equation of the normal at $(0, 1)$ is $y - 1 = -\frac{1}{2}(x - 0)$,which simplifies to $x + 2y - 2 = 0$.To find the intercepts with the axes:For the $x$-intercept,set $y = 0$: $x + 2(0) - 2 = 0 \implies x = 2$. The point is $(2, 0)$.For the $y$-intercept,set $x = 0$: $0 + 2y - 2 = 0 \implies y = 1$. The point is $(0, 1)$.The triangle formed by the normal and the axes has vertices at $(0, 0)$,$(2, 0)$,and $(0, 1)$.The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 1 = 1$ square unit.

The area of the triangle formed by the normal to the curve $y = e^{2x} + x^2$ at the point $(0, 1)$ with the coordinate axes is $......$ square units.

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