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. \(\sqrt{\frac{15}{2}}\)
• D. \(\sqrt{\frac{19}{2}}\)

Hint:

To find the minimum distance from the origin to a curve, minimize \[ D^2=x^2+y^2 \] instead of \(D\), because both have the same minimum point and \(D^2\) is easier to differentiate.

Answer 1

The Correct Option is A

Step 1: Set up the distance squared.
Point on the curve: \((x, x^2-4)\). \(D^2 = x^2+(x^2-4)^2 = x^4-7x^2+16\).

Step 2: Minimise by substituting \(u=x^2\).
\(g(u) = u^2-7u+16\). \(g'(u)=2u-7=0 \Rightarrow u=7/2\). \(D^2_{\min} = \dfrac{49}{4}-\dfrac{49}{2}+16 = \dfrac{49-98+64}{4} = \dfrac{15}{4}\).

Step 3: Find minimum distance.
\(D_{\min} = \dfrac{\sqrt{15}}{2}\).
\[ \boxed{\dfrac{\sqrt{15}}{2}} \]