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

Answer: (A)

$\frac{28}{3}$

Answer 2

The area of a circle is given by $A = \pi r^2$.

1. For circle $C_1$, we are given that $A = 36\pi$. Therefore, the radius $r_1$ of $C_1$ is: $\pi r_1^2 = 36\pi \implies r_1 = 6.$
2. 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 $\frac{28}{3}$.