Question

Let \( f(x) = x^2 + 9 \), \( g(x) = \frac{x}{x-9} \), and \[ a = f \circ g(10), \, b = g \circ f(3). \]
If \( e \) and \( l \) denote the eccentricity and the length of the latus rectum of the ellipse \[ \frac{x^2}{a} + \frac{y^2}{b} = 1, \] then \( 8e^2 + l^2 \) is equal to:

• A. 16
• B. 8
• C. 6
• D. 12

Answer 1

The Correct Option is B

Step 1: Compute \(a = f(g(10))\)
Given g(x) = \(\frac{x}{x-9}\), compute g(10):

\[ g(10) = \frac{10}{10 - 9} = \frac{10}{1} = 10. \]

Now, substitute g(10) into \(f(x) = x^2 + 9:\)

\[ f(g(10)) = f(10) = 10^2 + 9 = 100 + 9 = 109. \]

Thus: a = 109.

Step 2: Compute \(b = g(f(3))\)
Given \(f(x) = x^2 + 9\), compute f(3):

\[ f(3) = 3^2 + 9 = 9 + 9 = 18. \]

Now, substitute f(3) into \(g(x) = \frac{x}{x-9}:\)

\[ g(f(3)) = g(18) = \frac{18}{18 - 9} = \frac{18}{9} = 2. \]

Thus: b = 2.

Step 3: Equation of the ellipse The equation of the ellipse is:

\[ \frac{x^2}{a} + \frac{y^2}{b} = 1. \]

Substitute a = 109 and b = 2:

\[ \frac{x^2}{109} + \frac{y^2}{2} = 1. \]

Step 4: Eccentricity e The eccentricity of an ellipse is given by:

\[ e^2 = 1 - \frac{\text{smaller denominator}}{\text{larger denominator}}. \]

Here, the larger denominator is a = 109, and the smaller denominator is b = 2:

\[ e^2 = 1 - \frac{2}{109}. \]

\[ e^2 = \frac{109}{109} - \frac{2}{109} = \frac{107}{109}. \]

Step 5: Length of the latus rectum \(\ell\) The length of the latus rectum of an ellipse is given by:

\[ \ell = \frac{2b}{\sqrt{a}}. \]

Substitute b = 2 and a = 109:

\[ \ell = \frac{2(2)}{\sqrt{109}} = \frac{4}{\sqrt{109}}. \]

Step 6: Compute \(8e^2 + \ell^2\) First, compute \(\ell^2:\)

\[ \ell^2 = \left(\frac{4}{\sqrt{109}}\right)^2 = \frac{16}{109}. \]

Now compute \(8e^2:\)

\[ 8e^2 = 8 \times \frac{107}{109} = \frac{856}{109}. \]

Finally:

\[ 8e^2 + \ell^2 = \frac{856}{109} + \frac{16}{109} = \frac{872}{109} = 8. \]

Final Answer: Option (2).