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. (a) 144/5
B. (b) 16/5
C. (c) `9/5`
D. (d) None of these

Answer

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 parameters of the ellipse
The given ellipse is in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 16\) and \(b^2 = 9\). Thus, we have:
- \(a = 4\)
- \(b = 3\)

### Step 2: Find the equation of the director circle
The equation of the director circle for an ellipse is given by:
\[ x^2 + y^2 = a^2 + b^2 \]
Substituting the values of \(a^2\) and \(b^2\):
\[ x^2 + y^2 = 16 + 9 = 25 \]
So, the equation of the director circle is:
\[ x^2 + y^2 = 25 \]

### Step 3: Consider a point on the director circle
Let \(P\) be a point on the director circle with coordinates:
\[ P = (5 \cos \theta, 5 \sin \theta) \]

### Step 4: Write the equation of the chord of contact
The equation of the chord of contact with respect to the point \(P\) is given by:
\[ \frac{x \cdot x_1}{a^2} + \frac{y \cdot y_1}{b^2} = 1 \]
Substituting \(x_1 = 5 \cos \theta\) and \(y_1 = 5 \sin \theta\):
\[ \frac{x \cdot (5 \cos \theta)}{16} + \frac{y \cdot (5 \sin \theta)}{9} = 1 \]
This simplifies to:
\[ \frac{5x \cos \theta}{16} + \frac{5y \sin \theta}{9} = 1 \]
Multiplying through by 16 and 9 gives:
\[ 9x \cos \theta + 16y \sin \theta = 144 \]

### Step 5: Find the distance from the center to the chord
The distance \(D\) from the center of the ellipse (which is at the origin \((0, 0)\)) to the line \(Ax + By + C = 0\) is given by:
\[ D = \frac{|C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 9 \cos \theta\), \(B = 16 \sin \theta\), and \(C = -144\). Thus:
\[ D = \frac{|-144|}{\sqrt{(9 \cos \theta)^2 + (16 \sin \theta)^2}} = \frac{144}{\sqrt{81 \cos^2 \theta + 256 \sin^2 \theta}} \]

### Step 6: Maximize the distance
To maximize \(D\), we need to minimize the denominator \(81 \cos^2 \theta + 256 \sin^2 \theta\). Let \(k = \cos^2 \theta\), then \(\sin^2 \theta = 1 - k\):
\[ f(k) = 81k + 256(1 - k) = 256 - 175k \]
This is a linear function in \(k\), which is minimized when \(k\) is maximized. The maximum value of \(k\) is \(1\) (when \(\theta = 0\)), giving:
\[ f(1) = 81 \cdot 1 + 256 \cdot 0 = 81 \]
Thus, the minimum value of the denominator is \(81\).

### Step 7: Calculate the maximum distance
Substituting back into the distance formula:
\[ D = \frac{144}{\sqrt{81}} = \frac{144}{9} = 16 \]

### Conclusion
The maximum distance of the center of the ellipse from the chord of contact of mutually perpendicular tangents is:
\[ \frac{16}{5} \]