An equation of the tangent to the curve y = x | vedclass

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(A) Let the point of contact be $(h, k)$,where $k = h^4$. The slope of the tangent to the curve $y = x^4$ at $(h, k)$ is given by $\frac{dy}{dx} = 4x^

Vedclass · Accepted Answer

$y = 0$

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(A) Let the point of contact be $(h, k)$,where $k = h^4$. The slope of the tangent to the curve $y = x^4$ at $(h, k)$ is given by $\frac{dy}{dx} = 4x^3$. At $x = h$,the slope is $m = 4h^3$. The equation of the tangent at $(h, k)$ is $y - k = 4h^3(x - h)$. Since the tangent passes through the point $(2, 0)$,we substitute these coordinates into the equation: $0 - h^4 = 4h^3(2 - h)$ $-h^4 = 8h^3 - 4h^4$ $3h^4 - 8h^3 = 0$ $h^3(3h - 8) = 0$ This gives $h = 0$ or $h = \frac{8}{3}$. For $h = 0$,the equation of the tangent is $y - 0 = 4(0)^3(x - 0)$,which simplifies to $y = 0$. For $h = \frac{8}{3}$,the equation of the tangent is $y - (\frac{8}{3})^4 = 4(\frac{8}{3})^3(x - \frac{8}{3})$. Thus,$y = 0$ is one of the equations of the tangent.

An equation of the tangent to the curve $y = x^4$ from the point $(2, 0)$ not on the curve is

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