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The point on the curve $y=x^3$, at which the tangent to the curve is parallel to the $X$-axis, is
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Verified Answer
The correct answer is:
$(0,0)$
Given,
curve, $y=x^3$
$\because$ Slope of tangent to the curve at any point $\left(x_1, y_1\right)$ on the curve is $=\left.\frac{d y}{d x}\right|_{\left(x_1, y_1\right)}$
So, $\frac{d y}{d x}=3 x^2$
$\therefore$ Slope of tangent at $\left(x_1, y_1\right)$ is $m_T=3 x_1^2$
$\because$ Tangent is parallel to $X$-axis $\Rightarrow m_T=$ Slope of $X$-axis
$$
\begin{array}{ll}
\Rightarrow & m_T=0 \Rightarrow 3 x_1^2=0 \\
\Rightarrow & x_1=0 \Rightarrow y_1=0
\end{array}
$$
$\therefore$ Point of contact is $(0,0)$
curve, $y=x^3$
$\because$ Slope of tangent to the curve at any point $\left(x_1, y_1\right)$ on the curve is $=\left.\frac{d y}{d x}\right|_{\left(x_1, y_1\right)}$
So, $\frac{d y}{d x}=3 x^2$
$\therefore$ Slope of tangent at $\left(x_1, y_1\right)$ is $m_T=3 x_1^2$
$\because$ Tangent is parallel to $X$-axis $\Rightarrow m_T=$ Slope of $X$-axis
$$
\begin{array}{ll}
\Rightarrow & m_T=0 \Rightarrow 3 x_1^2=0 \\
\Rightarrow & x_1=0 \Rightarrow y_1=0
\end{array}
$$
$\therefore$ Point of contact is $(0,0)$
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