Question

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

• A. $\dfrac{\sqrt{15}}{2}$
• B. $\sqrt{\dfrac{19}{2}}$
• C. $\sqrt{\dfrac{15}{2}}$
• D. $\dfrac{\sqrt{19}}{2}$

Hint:

To find minimum distance from a curve to a point, minimize the square of the distance instead of the distance itself.

Answer 1

The Correct Option is B

Step 1: Let a point on the curve be $(x,\,y)$ where \[ y=x^2-4 \]
Step 2: Distance of this point from the origin is: \[ D=\sqrt{x^2+y^2} \] To minimize $D$, minimize $D^2$: \[ D^2=x^2+(x^2-4)^2 \]
Step 3: Simplify: \[ D^2=x^2+x^4-8x^2+16 = x^4-7x^2+16 \]
Step 4: Differentiate with respect to $x$: \[ \frac{d(D^2)}{dx}=4x^3-14x \] Set derivative equal to zero: \[ 4x^3-14x=0 \Rightarrow 2x(2x^2-7)=0 \]
Step 5: Critical points: \[ x=0,\quad x=\pm\sqrt{\frac{7}{2}} \]
Step 6: Evaluate $D^2$ at these points: For $x=0$: \[ D^2=16 \] For $x^2=\frac{7}{2}$: \[ D^2=\left(\frac{7}{2}\right)^2-7\left(\frac{7}{2}\right)+16 =\frac{49}{4}-\frac{49}{2}+16 =\frac{19}{2} \]
Step 7: Minimum distance: \[ D=\sqrt{\frac{19}{2}} \]