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

Answer: (B)

8

Answer 2

**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).