Question

If the radius of the largest circle with centre (2, 0) inscribed in the ellipse \(x^2 + 4y^2 = 36\) is r, then \(12r^2\) is equal to

• A. 92
• B. 69
• C. 72
• D. 115

Answer 1

The Correct Option is A

Given:

• Center of the ellipse: \( C(2, 0) \) 
• Equation of the ellipse: \[ \frac{x^2}{36} + \frac{y^2}{9} = 1 \]

Step 1: Equation of the Normal at a Point \( P \) on the Ellipse

The parametric coordinates of a point \( P \) on the ellipse are given by:

\[ P(6\cos\theta, 3\sin\theta). \]

The equation of the normal at \( P \) is:

\[ (6\sec\theta)x - (3\csc\theta)y = 27. \]

Step 2: Normal Passes Through \( C(2, 0) \)

Substitute \( C(2, 0) \) into the normal equation:

\[ (6\sec\theta)(2) - (3\csc\theta)(0) = 27. \]

Simplify:

\[ 12\sec\theta = 27 \quad \Rightarrow \quad \sec\theta = \frac{27}{12} = \frac{9}{4}. \]

Step 3: Determine \( \cos\theta \) and \( \sin\theta \)

Using \( \sec\theta = \frac{1}{\cos\theta} \):

\[ \cos\theta = \frac{4}{9}. \]

Find \( \sin\theta \) using \( \sin^2\theta = 1 - \cos^2\theta \):

\[ \sin\theta = \sqrt{1 - \left(\frac{4}{9}\right)^2} = \sqrt{\frac{81 - 16}{81}} = \sqrt{\frac{65}{81}} = \frac{\sqrt{65}}{9}. \]

Step 4: Parametric Point \( P \)

The parametric coordinates of \( P \) are:

\[ P = \left(6\cos\theta, 3\sin\theta\right) = \left(6 \cdot \frac{4}{9}, 3 \cdot \frac{\sqrt{65}}{9}\right) = \left(\frac{8}{3}, \frac{\sqrt{65}}{3}\right). \]

Step 5: Distance Between \( P \) and \( C \)

The distance \( \gamma \) between \( P\left(\frac{8}{3}, \frac{\sqrt{65}}{3}\right) \) and \( C(2, 0) \) is:

\[ \gamma = \sqrt{\left(\frac{8}{3} - 2\right)^2 + \left(\frac{\sqrt{65}}{3} - 0\right)^2}. \]

Simplify the terms:

\[ \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3}. \]

Thus:

\[ \gamma = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{\sqrt{65}}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{65}{9}} = \sqrt{\frac{69}{9}} = \frac{\sqrt{69}}{3}. \]

Step 6: Value of \( 12\gamma^2 \)

The value of \( 12\gamma^2 \) is:

\[ 12\gamma^2 = 12 \cdot \left(\frac{\sqrt{69}}{3}\right)^2 = 12 \cdot \frac{69}{9} = \frac{828}{9} = 92. \]

Final Answer:

The value of \( 12\gamma^2 \) is \( 92 \).