For an ellipse having major and minor axis along `x` and `y` axes respectivley, the product of semi major and semi minor axis is `20`. Then maximum value of product of abscissa and ordinate of any point on the ellipse is greater than (A) `5` (B) `8` (C) `10` (D) `15`

Options

A. `5`
B. `8`
C. `10`
D. `15`

Explanation

To solve the problem, we need to find the maximum value of the product of the abscissa (x-coordinate) and ordinate (y-coordinate) of any point on the ellipse defined by the given conditions.

### Step-by-step Solution:

1. **Understanding the Ellipse Equation**:
The general equation of an ellipse with the major axis along the x-axis and the minor axis along the y-axis is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

2. **Given Condition**:
We are given that the product of the semi-major and semi-minor axes is:
\[ ab = 20 \]

3. **Finding the Maximum Product \( P = xy \)**:
We want to maximize the product \( P = xy \). To express \( y \) in terms of \( x \), we can rearrange the ellipse equation:
\[ y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right) \]
Thus,
\[ y = b \sqrt{1 - \frac{x^2}{a^2}} \]
Therefore, the product \( P \) can be expressed as:
\[ P = x \cdot b \sqrt{1 - \frac{x^2}{a^2}} = b \cdot x \sqrt{1 - \frac{x^2}{a^2}} \]

4. **Using AM-GM Inequality**:
To find the maximum value of \( P \), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. We know that:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \]
This implies:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
By AM-GM, we have:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 2 \sqrt{\frac{x^2}{a^2} \cdot \frac{y^2}{b^2}} \]
Setting \( u = \frac{x^2}{a^2} \) and \( v = \frac{y^2}{b^2} \), we get:
\[ 1 \geq 2 \sqrt{uv} \]
Thus:
\[ \frac{1}{4} \geq uv \implies uv \leq \frac{1}{4} \]

5. **Substituting Back**:
Since \( u = \frac{x^2}{a^2} \) and \( v = \frac{y^2}{b^2} \), we have:
\[ \frac{x^2 y^2}{a^2 b^2} \leq \frac{1}{4} \implies x^2 y^2 \leq \frac{a^2 b^2}{4} \]
Taking the square root gives:
\[ xy \leq \frac{ab}{2} \]

6. **Substituting the Value of \( ab \)**:
Given \( ab = 20 \):
\[ xy \leq \frac{20}{2} = 10 \]

7. **Conclusion**:
The maximum value of the product \( xy \) is \( 10 \). Therefore, the maximum value of the product of the abscissa and ordinate of any point on the ellipse is greater than \( 10 \).

### Answer:
The maximum value of the product of abscissa and ordinate of any point on the ellipse is greater than \( 10 \). Thus, the correct option is:
(C) \( 10 \).