The minimum distance of a point on the curve | vedclass

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Question

(A) Let a point on the curve be $P(\alpha, \alpha^2 - 4)$. The distance $D$ of this point from the origin $(0,0)$ is given by $D = \sqrt{\alpha^2 + (\

Vedclass · Accepted Answer

$\frac{\sqrt{15}}{2}$

Vedclass · Answer

(A) Let a point on the curve be $P(\alpha, \alpha^2 - 4)$. The distance $D$ of this point from the origin $(0,0)$ is given by $D = \sqrt{\alpha^2 + (\alpha^2 - 4)^2}$.To minimize $D$,we minimize $f(\alpha) = D^2 = \alpha^2 + (\alpha^2 - 4)^2$.$f(\alpha) = \alpha^2 + \alpha^4 - 8\alpha^2 + 16 = \alpha^4 - 7\alpha^2 + 16$.Taking the derivative with respect to $\alpha$ and setting it to zero:$f'(\alpha) = 4\alpha^3 - 14\alpha = 0$.$2\alpha(2\alpha^2 - 7) = 0$.This gives $\alpha = 0$ or $\alpha^2 = \frac{7}{2}$.If $\alpha^2 = 0$,$f(0) = 16$.If $\alpha^2 = \frac{7}{2}$,$f(\alpha) = (\frac{7}{2})^2 - 7(\frac{7}{2}) + 16 = \frac{49}{4} - \frac{49}{2} + 16 = -\frac{49}{4} + 16 = \frac{15}{4}$.Since $\frac{15}{4} Thus,the minimum distance $D = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}$.

The minimum distance of a point on the curve $y = x^2 - 4$ from the origin is

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