The area of the triangle formed by the coordi | vedclass

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(B) Given the curve $y = e^{2x} + x^2$. At $x = 0$,$y = e^0 + 0^2 = 1 + 0 = 1$. So,the point of contact is $(0, 1)$. Now,find the derivative $\frac{dy

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(B) Given the curve $y = e^{2x} + x^2$. At $x = 0$,$y = e^0 + 0^2 = 1 + 0 = 1$. So,the point of contact is $(0, 1)$. Now,find the derivative $\frac{dy}{dx} = 2e^{2x} + 2x$. At $x = 0$,the slope of the tangent $m_t = 2e^0 + 2(0) = 2$. The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{2}$. The equation of the normal at $(0, 1)$ is $y - 1 = -\frac{1}{2}(x - 0)$,which simplifies to $y = -\frac{1}{2}x + 1$ or $x + 2y = 2$. The normal intersects the $x$-axis at $y = 0$,so $x + 2(0) = 2 \implies x = 2$. The point is $(2, 0)$. The normal intersects the $y$-axis at $x = 0$,so $0 + 2y = 2 \implies y = 1$. The point is $(0, 1)$. The area of the triangle formed by the axes and the normal is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 1 = 1$.

The area of the triangle formed by the coordinate axes and the normal to the curve $y = e^{2x} + x^2$ at the point $x = 0$.

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