$\begin{array}{1 1} (A)\;\text{a vertical tangent (parallel to y-axis)}\\(B)\;\text{a horizontal tangent (parallel to x-axis)}\\(C)\;\text{an oblique tangent}\\(D)\;\text{no tangent}\end{array} $

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- Slope of a line is $ ax+by+c=0$ is $ - \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
- If $ y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p = $ slope of the tangent to $ y = f(x)$ at point $p$

Step 1

The equation of the curve is

$ y=x^{\large\frac{1}{5}}$

differentiting w.r.t $x$ we get,

$ \large\frac{dy}{dx} = \large\frac{1}{5}x^{\large\frac{-4}{5}}=\large\frac{1}{5x^{\large\frac{4}{5}}}$

The point given is $(0, 0)$

$ \Rightarrow \large\frac{dy}{dx_{(0,0)}}=\large\frac{1}{0}$$=\infty$

The slope of the tangent is $\infty$

$\Rightarrow\:$ The tangent is parallel to $y$ - axis

Hence the correct option is A

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