The points on the curve y=frac{2}{3} x^3+frac | vedclass

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(A) Given curve is $y=\frac{2}{3} x^3+\frac{1}{2} x^2$. Differentiating with respect to $x$,we get $\frac{d y}{d x} = 2x^2 + x$. Since the tangents ma

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$\left(\frac{1}{2}, \frac{5}{24}\right)$ and $\left(-1, \frac{-1}{6}\right)$

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(A) Given curve is $y=\frac{2}{3} x^3+\frac{1}{2} x^2$. Differentiating with respect to $x$,we get $\frac{d y}{d x} = 2x^2 + x$. Since the tangents make equal angles with the coordinate axes,the slope of the tangent must be $\pm 1$. Case $1$: $\frac{d y}{d x} = 1 \Rightarrow 2x^2 + x = 1 \Rightarrow 2x^2 + x - 1 = 0$. Solving this quadratic equation: $(2x - 1)(x + 1) = 0$,so $x = \frac{1}{2}$ or $x = -1$. For $x = \frac{1}{2}$,$y = \frac{2}{3}(\frac{1}{8}) + \frac{1}{2}(\frac{1}{4}) = \frac{1}{12} + \frac{1}{8} = \frac{5}{24}$. For $x = -1$,$y = \frac{2}{3}(-1) + \frac{1}{2}(1) = -\frac{2}{3} + \frac{1}{2} = -\frac{1}{6}$. Case $2$: $\frac{d y}{d x} = -1 \Rightarrow 2x^2 + x = -1 \Rightarrow 2x^2 + x + 1 = 0$. The discriminant $D = 1^2 - 4(2)(1) = 1 - 8 = -7 Thus,the required points are $\left(\frac{1}{2}, \frac{5}{24}\right)$ and $\left(-1, \frac{-1}{6}\right)$.

The points on the curve $y=\frac{2}{3} x^3+\frac{1}{2} x^2$,where the tangents make equal angles with coordinate axes are

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