The maximum distance of the centre of the ellipse `(x^(2))/(16) +(y^(2))/(9) =1` from the chord of contact of mutually perpendicular tangents of the ellipse is

Options

A. `(144)/(5)`
B. `(9)/(5)`
C. `(16)/(5)`
D. `(8)/(5)`

Explanation

To solve the problem of finding the maximum distance of the center of the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) from the chord of contact of mutually perpendicular tangents, we can follow these steps:

### Step 1: Identify the center of the ellipse
The given ellipse is centered at the origin \((0, 0)\).

### Step 2: Determine the equation of the director circle
The director circle of an ellipse is given by the equation:
\[ x^2 + y^2 = a^2 + b^2 \]
where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. For our ellipse, \(a^2 = 16\) and \(b^2 = 9\). Thus,
\[ a = 4 \quad \text{and} \quad b = 3 \]
Calculating \(a^2 + b^2\):
\[ a^2 + b^2 = 16 + 9 = 25 \]
Therefore, the equation of the director circle is:
\[ x^2 + y^2 = 25 \]

### Step 3: Parametrize a point on the director circle
Let \(P\) be a point on the director circle. We can express the coordinates of \(P\) using the parameter \(\theta\):
\[ P = (5 \cos \theta, 5 \sin \theta) \]

### Step 4: Write the equation of the chord of contact
The chord of contact for the ellipse with respect to point \(P(x_1, y_1)\) is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Substituting \(x_1 = 5 \cos \theta\) and \(y_1 = 5 \sin \theta\):
\[ \frac{x(5 \cos \theta)}{16} + \frac{y(5 \sin \theta)}{9} = 1 \]
This simplifies to:
\[ \frac{5x \cos \theta}{16} + \frac{5y \sin \theta}{9} = 1 \]
Dividing through by 5 gives:
\[ \frac{x \cos \theta}{16} + \frac{y \sin \theta}{9} = \frac{1}{5} \]

### Step 5: Calculate the distance from the center to the chord of contact
The distance \(D\) from the center \((0, 0)\) to the line \(Ax + By + C = 0\) is given by:
\[ D = \frac{|C|}{\sqrt{A^2 + B^2}} \]
Rearranging our chord equation:
\[ \frac{x \cos \theta}{16} + \frac{y \sin \theta}{9} - \frac{1}{5} = 0 \]
Here, \(A = \frac{\cos \theta}{16}\), \(B = \frac{\sin \theta}{9}\), and \(C = -\frac{1}{5}\). Thus, the distance \(D\) becomes:
\[ D = \frac{\left| -\frac{1}{5} \right|}{\sqrt{\left(\frac{\cos \theta}{16}\right)^2 + \left(\frac{\sin \theta}{9}\right)^2}} \]
This simplifies to:
\[ D = \frac{\frac{1}{5}}{\sqrt{\frac{\cos^2 \theta}{256} + \frac{\sin^2 \theta}{81}}} \]

### Step 6: Maximize the distance
To find the maximum distance, we need to maximize:
\[ D = \frac{1}{5} \cdot \frac{1}{\sqrt{\frac{\cos^2 \theta}{256} + \frac{\sin^2 \theta}{81}}} \]
Let \(x = \cos^2 \theta\) and \(y = \sin^2 \theta\) such that \(x + y = 1\). We can express the denominator as:
\[ D = \frac{1}{5 \sqrt{\frac{x}{256} + \frac{1-x}{81}}} \]

### Step 7: Solve for maximum distance
To find the maximum distance, we can set \(\cos \theta = 1\) and \(\sin \theta = 0\) (which gives the maximum value):
\[ D = \frac{1}{5 \sqrt{\frac{1}{256}}} = \frac{1}{5} \cdot 16 = \frac{16}{5} \]

### Final Answer
Thus, the maximum distance of the center of the ellipse from the chord of contact of mutually perpendicular tangents is:
\[ \boxed{\frac{16}{5}} \]