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The slope of normal to the curve $y=3x^2-6x$ at x= 0 is : |
-6 $-\frac{1}{6}$ $\frac{1}{6}$ $-\frac{2}{3}$ |
$\frac{1}{6}$ |
$y=3x^2-6x$ differentiating wrt x $\frac{dy}{dx}=6x-6$ (Slope of tangent) $\frac{-dx}{dy}=\frac{1}{6-6x}$ (Slope of Normal) at $x=6$ → $\frac{-dx}{dy}]_{x=0}=\frac{1}{6}$ |
