Question

Consider a hyperbola \( H \) having its centre at the origin and foci on the \( x \)-axis. Let \( C_1 \) be the circle touching the hyperbola \( H \) and having its centre at the origin. Let \( C_2 \) be the circle touching the hyperbola \( H \) at its vertex and having its centre at one of its foci. If the areas (in square units) of \( C_1 \) and \( C_2 \) are \( 36\pi \) and \( 4\pi \), respectively, then the length (in units) of the latus rectum of \( H \) is:

• A. \( \frac{28}{3} \)
• B. \( \frac{14}{3} \)
• C. \( \frac{10}{3} \)
• D. \( \frac{11}{3} \)

Answer 1

The Correct Option is A

The area of a circle is computed using the formula \( A = \pi r^2 \).

1. For circle \( C_1 \), the given area is \( A = 36\pi \). This implies the radius \( r_1 \) of \( C_1 \) is calculated as: \[ \pi r_1^2 = 36\pi \implies r_1 = 6. \]
2. For circle \( C_2 \), the given area is \( A = 4\pi \). This implies the radius \( r_2 \) of \( C_2 \) is calculated as: \[ \pi r_2^2 = 4\pi \implies r_2 = 2. \]

Given that \( C_1 \) is centered at the origin and is tangent to the hyperbola, its radius \( r_1 = 6 \) equals the distance from the center to the hyperbola's vertex, which is denoted by \( a \) (the semi-major axis length). Consequently, \( a = 6 \).

For a hyperbola centered at the origin with foci on the \( x \)-axis, the focal distance \( c \) is the distance from the origin to a focus. Since \( C_2 \) is centered at one of the foci and has a radius \( r_2 = 2 \), the relationship \( c - a = 2 \) holds. Therefore, \[ c = a + 2 = 6 + 2 = 8. \]

The relationship among \( a \), \( b \), and \( c \) for a hyperbola is \( c^2 = a^2 + b^2 \). Substituting the known values yields: \[ 8^2 = 6^2 + b^2 \implies 64 = 36 + b^2 \implies b^2 = 28 \implies b = \sqrt{28}. \]

The length of the latus rectum for a hyperbola is calculated by \( \frac{2b^2}{a} \). Thus, the latus rectum length is: \[ \frac{2b^2}{a} = \frac{2 \times 28}{6} = \frac{56}{6} = \frac{28}{3}. \]

The length of the latus rectum of \( H \) is determined to be \( \frac{28}{3} \).