The distance between the origin and the norma | vedclass

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(C) Given curve is $y = e^{2x} + x^2$. At $x = 0$,$y = e^0 + 0^2 = 1$. So,the point of contact is $(0, 1)$. Differentiating with respect to $x$,we get

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$\frac{2}{\sqrt{5}}$

Vedclass · Answer

(C) Given curve is $y = e^{2x} + x^2$. At $x = 0$,$y = e^0 + 0^2 = 1$. So,the point of contact is $(0, 1)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2e^{2x} + 2x$. 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 $y - 1 = -\frac{1}{2}(x - 0)$. $2y - 2 = -x$,which simplifies to $x + 2y - 2 = 0$. The distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$. Here,$A = 1$,$B = 2$,and $C = -2$. $d = \frac{|-2|}{\sqrt{1^2 + 2^2}} = \frac{2}{\sqrt{1 + 4}} = \frac{2}{\sqrt{5}}$ units.

The distance between the origin and the normal to the curve $y = e^{2x} + x^2$ drawn at $x = 0$ is . . . . . . units

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