If the tangent at a point P on the curve y=4x | vedclass

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Question

(A) Given the curve $y=4x^4+x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 16x^3+1$. The slope of the tangent at $(0,0)$ is $m_1 = 16

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$\left(\frac{-1}{2}, \frac{-1}{4}\right)$

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(A) Given the curve $y=4x^4+x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 16x^3+1$. The slope of the tangent at $(0,0)$ is $m_1 = 16(0)^3+1 = 1$. Let the point $P$ be $(a, b)$. The slope of the tangent at $P$ is $m_2 = 16a^3+1$. Since the tangents are perpendicular,$m_1 	imes m_2 = -1$. $1 	imes (16a^3+1) = -1$. $16a^3 = -2$ $\Rightarrow a^3 = -\frac{1}{8}$ $\Rightarrow a = -\frac{1}{2}$. Since $P$ lies on the curve,$b = 4(-\frac{1}{2})^4 + (-\frac{1}{2}) = 4(\frac{1}{16}) - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$. Thus,the point $P$ is $\left(-\frac{1}{2}, -\frac{1}{4}ight)$.

If the tangent at a point $P$ on the curve $y=4x^4+x$ is perpendicular to the tangent to the same curve at $(0,0)$,then the point $P$ is

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