If the least area of triangle formed by tangent,normal at any point P on the curve `y = f(x)` and X-axis is 4 sq. unit. Then the ordinate of the point P (P lies in first quadrant) is

Options

A. 1
B. `(1)/(2)`
C. `(1)/(4)`
D. `2`

Answer

To solve the problem, we need to find the ordinate (y-coordinate) of the point P on the curve \( y = f(x) \) such that the least area of the triangle formed by the tangent, normal at point P, and the X-axis is 4 square units.

### Step-by-Step Solution:

1. **Understanding the Triangle Formation**:
- At point \( P(x, y) \) on the curve \( y = f(x) \), we draw the tangent and normal lines.
- The triangle formed by these lines and the X-axis will have its vertices at points \( P \), where the tangent intersects the X-axis (let's call this point \( A \)), and where the normal intersects the X-axis (let's call this point \( B \)).

2. **Area of the Triangle**:
- The area \( A \) of triangle \( APB \) can be expressed as:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
- The base of the triangle is the distance between points \( A \) and \( B \) on the X-axis, and the height is the ordinate \( y \) of point \( P \).

3. **Finding the Lengths**:
- The tangent line at point \( P(x, y) \) has a slope of \( f'(x) \), and the equation of the tangent line is:
\[ y - f(x) = f'(x)(x - x_0) \]
- The normal line has a slope of \( -\frac{1}{f'(x)} \) and its equation is:
\[ y - f(x) = -\frac{1}{f'(x)}(x - x_0) \]

4. **Calculating the Area**:
- The area can be expressed in terms of \( y \) and the angles formed by the tangent and normal lines. The area \( A \) can be simplified to:
\[ A = \frac{1}{2} y^2 (\tan \theta + \cot \theta) \]
- We need to minimize this area.

5. **Using Inequalities**:
- By applying the AM-GM inequality:
\[ \tan \theta + \cot \theta \geq 2 \]
- Hence, the minimum value of \( \tan \theta + \cot \theta \) is 2.

6. **Finding Minimum Area**:
- Substituting this back into the area formula gives:
\[ A_{\text{min}} = \frac{1}{2} y^2 \cdot 2 = y^2 \]
- We are given that the least area is 4 square units, so:
\[ y^2 = 4 \]
- Thus, \( y = \pm 2 \).

7. **Considering the First Quadrant**:
- Since point \( P \) lies in the first quadrant, we take the positive value:
\[ y = 2 \]

### Conclusion:
The ordinate of the point \( P \) is \( 2 \).