Question

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

• A. \( \frac{\sqrt{15}}{2} \)
• B. \( \frac{\sqrt{19}}{2} \)
• C. \( \sqrt{\frac{15}{2}} \)
  % Corrected from previous image transcription, this seems more plausible.
• D. \( \sqrt{\frac{19}{2}} \)

Hint:

To find the minimum/maximum distance between a point and a curve, it's often easier to minimize/maximize the square of the distance.
Let the point on the curve be \((x, f(x))\). Distance squared from origin is \(S = x^2 + (f(x))^2\).
Find critical points by setting \(dS/dx = 0\).
Use the second derivative test (\(d^2S/dx^2\)) to classify critical points (minimum if \(S'>0\), maximum if \(S''<0\)).

Answer 1

The Correct Option is A

Let P\((x,y)\) be a point on the curve \(y = x^2 - 4\). The distance \(D\) of P from the origin O(0,0) is given by \(D = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2+y^2}\). To minimize D, we can minimize \(D^2\). Let \(S = D^2 = x^2+y^2\). Since \(y = x^2-4\), substitute this into the expression for S: \(S(x) = x^2 + (x^2-4)^2\) \(S(x) = x^2 + (x^4 - 8x^2 + 16)\) \(S(x) = x^4 - 7x^2 + 16\). To find the minimum value of S, we find the derivative \(S'(x)\) and set it to zero. \(S'(x) = \frac{dS}{dx} = 4x^3 - 14x\). Set \(S'(x) = 0\): \(4x^3 - 14x = 0\) \(2x(2x^2 - 7) = 0\). This gives possible values for x: 1. \(x = 0\) 2. \(2x^2 - 7 = 0 \Rightarrow 2x^2 = 7 \Rightarrow x^2 = 7/2 \Rightarrow x = \pm\sqrt{7/2}\). Now we find \(S''(x)\) to check for minima/maxima. \(S''(x) = \frac{d^2S}{dx^2} = 12x^2 - 14\). Evaluate \(S''(x)\) at the critical points: \begin{itemize} \item At \(x=0\): \(S''(0) = 12(0)^2 - 14 = -14\). Since \(S''(0)<0\), this corresponds to a local maximum for S. \item At \(x^2 = 7/2\): \(S''(x) = 12(7/2) - 14 = 6 \times 7 - 14 = 42 - 14 = 28\). Since \(S''(x)>0\), these points (\(x=\pm\sqrt{7/2}\)) correspond to local minima for S. \end{itemize} The minimum value of S occurs when \(x^2 = 7/2\). Substitute \(x^2 = 7/2\) into the expression for S: \(S_{min} = (x^2)^2 - 7x^2 + 16 = (7/2)^2 - 7(7/2) + 16\) \(S_{min} = \frac{49}{4} - \frac{49}{2} + 16 = \frac{49 - 98 + 64}{4} = \frac{113 - 98}{4} = \frac{15}{4}\). This is the minimum value of \(D^2\). The minimum distance \(D_{min}\) is \(\sqrt{S_{min}}\). \(D_{min} = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{\sqrt{4}} = \frac{\sqrt{15}}{2}\). This matches option (a). \[ \boxed{\frac{\sqrt{15}}{2}} \]