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:
- \( \frac{28}{3} \)
- \( \frac{14}{3} \)
- \( \frac{10}{3} \)
- \( \frac{11}{3} \)
The Correct Option is A
Solution and Explanation
The area of a circle is given by \( A = \pi r^2 \).
- For circle \( C_1 \), we are given that \( A = 36\pi \). Therefore, the radius \( r_1 \) of \( C_1 \) is: \[ \pi r_1^2 = 36\pi \implies r_1 = 6. \]
- For circle \( C_2 \), we are given that \( A = 4\pi \). Therefore, the radius \( r_2 \) of \( C_2 \) is: \[ \pi r_2^2 = 4\pi \implies r_2 = 2. \]
Since \( C_1 \) is centered at the origin and touches the hyperbola, the radius \( r_1 = 6 \) is equal to the distance from the center to the vertex of the hyperbola, which is \( a \) (the semi-major axis length). Therefore, \( a = 6 \).
For the hyperbola centered at the origin with foci along the \( x \)-axis, the focal distance \( c \) is the distance from the origin to one of the foci. Since \( C_2 \) has its center at one of the foci and radius \( r_2 = 2 \), we find that \( c - a = 2 \). Thus, \[ c = a + 2 = 6 + 2 = 8. \]
The relationship between \( a \), \( b \), and \( c \) for a hyperbola is given by \( c^2 = a^2 + b^2 \). Substituting the known values: \[ 8^2 = 6^2 + b^2 \implies 64 = 36 + b^2 \implies b^2 = 28 \implies b = \sqrt{28}. \]
The length of the latus rectum for a hyperbola is given by \( \frac{2b^2}{a} \). Therefore, the length of the latus rectum is: \[ \frac{2b^2}{a} = \frac{2 \times 28}{6} = \frac{56}{6} = \frac{28}{3}. \]
Thus, the length of the latus rectum of \( H \) is \( \frac{28}{3} \).
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