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

Options

A. `(sqrt(15))/(2)` units
B. `sqrt((19)/(2))` units
C. `sqrt((15)/(2))` units
D. `(sqrt(19))/(2)` units

Answer

To find the minimum distance of a point on the curve \( y = x^2 - 4 \) from the origin, we can follow these steps:

### Step 1: Define the Distance Formula
The distance \( D \) from a point \( (x, y) \) on the curve to the origin \( (0, 0) \) is given by the formula:
\[ D = \sqrt{x^2 + y^2} \]

### Step 2: Substitute the Curve Equation
Since the point lies on the curve \( y = x^2 - 4 \), we can substitute \( y \) in the distance formula:
\[ D = \sqrt{x^2 + (x^2 - 4)^2} \]

### Step 3: Simplify the Distance Formula
Now we simplify the expression:
\[ D = \sqrt{x^2 + (x^4 - 8x^2 + 16)} \]
\[ D = \sqrt{x^4 - 7x^2 + 16} \]

### Step 4: Minimize the Distance
To find the minimum distance, we can minimize \( D^2 \) instead of \( D \) (since the square root function is increasing). Let:
\[ f(x) = D^2 = x^4 - 7x^2 + 16 \]

### Step 5: Differentiate and Find Critical Points
Now, we differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = 4x^3 - 14x \]
Setting the derivative equal to zero to find critical points:
\[ 4x^3 - 14x = 0 \]
Factoring out \( 2x \):
\[ 2x(2x^2 - 7) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad 2x^2 - 7 = 0 \]
Solving \( 2x^2 - 7 = 0 \):
\[ 2x^2 = 7 \quad \Rightarrow \quad x^2 = \frac{7}{2} \quad \Rightarrow \quad x = \pm \sqrt{\frac{7}{2}} \]

### Step 6: Evaluate the Distance at Critical Points
We will evaluate \( D^2 \) at \( x = 0 \) and \( x = \sqrt{\frac{7}{2}} \):
1. For \( x = 0 \):
\[ D^2 = 0^4 - 7(0^2) + 16 = 16 \quad \Rightarrow \quad D = 4 \]
2. For \( x = \sqrt{\frac{7}{2}} \):
\[ D^2 = \left(\frac{7}{2}\right)^2 - 7\left(\frac{7}{2}\right) + 16 \]
\[ = \frac{49}{4} - \frac{49}{2} + 16 \]
\[ = \frac{49}{4} - \frac{98}{4} + \frac{64}{4} = \frac{49 - 98 + 64}{4} = \frac{15}{4} \quad \Rightarrow \quad D = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \]

### Step 7: Compare Distances
Now we compare the distances:
- At \( x = 0 \), \( D = 4 \)
- At \( x = \sqrt{\frac{7}{2}} \), \( D = \frac{\sqrt{15}}{2} \)

Since \( \frac{\sqrt{15}}{2} < 4 \), the minimum distance occurs at \( x = \sqrt{\frac{7}{2}} \).

### Final Result
Thus, the minimum distance from the origin to the curve \( y = x^2 - 4 \) is:
\[ \frac{\sqrt{15}}{2} \]