The minimum distance of a point on the curve | vedclass

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Question

(A) Let the point on the curve be $P(x, y) = (x, x^2-4)$.The square of the distance $D$ from the origin $(0,0)$ is $S = x^2 + y^2 = x^2 + (x^2-4)^2$.$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the point on the curve be $P(x, y) = (x, x^2-4)$.The square of the distance $D$ from the origin $(0,0)$ is $S = x^2 + y^2 = x^2 + (x^2-4)^2$.$S = x^2 + x^4 - 8x^2 + 16 = x^4 - 7x^2 + 16$.To find the minimum,differentiate $S$ with respect to $x$ and set it to zero:$\frac{dS}{dx} = 4x^3 - 14x = 0$.$2x(2x^2 - 7) = 0$.This gives $x = 0$ or $x^2 = \frac{7}{2}$.If $x = 0$,$S = 16$. If $x^2 = \frac{7}{2}$,$S = (\frac{7}{2})^2 - 7(\frac{7}{2}) + 16 = \frac{49}{4} - \frac{49}{2} + 16 = 16 - \frac{49}{4} = \frac{64-49}{4} = \frac{15}{4}$.The minimum distance is $\sqrt{S} = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2}$.

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

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