Let A be the area of the region in the first quadrant bounded by the x-axis the line 2y = x and the ellipse `(x^(2))/(9) + (y^(2))/(1) = 1`. Let A' be the area of the region in the first quadrant bounded by the y-axis the line y = kx and the ellipse . The value of 9k such that A and A' are equal is _____

Answer

To solve the problem, we need to find the areas \( A \) and \( A' \) and then determine the value of \( 9k \) such that \( A = A' \).

### Step 1: Determine the area \( A \)

The area \( A \) is bounded by the x-axis, the line \( 2y = x \) (which can be rewritten as \( y = \frac{1}{2}x \)), and the ellipse given by:

\[ \frac{x^2}{9} + \frac{y^2}{1} = 1 \]

First, we need to find the points of intersection between the line and the ellipse.

Substituting \( y = \frac{1}{2}x \) into the ellipse equation:

\[ \frac{x^2}{9} + \frac{(\frac{1}{2}x)^2}{1} = 1 \]

This simplifies to:

\[ \frac{x^2}{9} + \frac{x^2}{4} = 1 \]

To combine the fractions, we find a common denominator, which is 36:

\[ \frac{4x^2}{36} + \frac{9x^2}{36} = 1 \]

This gives:

\[ \frac{13x^2}{36} = 1 \implies 13x^2 = 36 \implies x^2 = \frac{36}{13} \implies x = \frac{6}{\sqrt{13}} \]

Now, substituting back to find \( y \):

\[ y = \frac{1}{2}x = \frac{1}{2} \cdot \frac{6}{\sqrt{13}} = \frac{3}{\sqrt{13}} \]

Thus, the points of intersection are \( \left(\frac{6}{\sqrt{13}}, \frac{3}{\sqrt{13}}\right) \).

### Step 2: Set up the integral for area \( A \)

The area \( A \) can be calculated using the integral:

\[ A = \int_0^{\frac{6}{\sqrt{13}}} \left( \frac{1}{2}x - 0 \right) \, dx \]

### Step 3: Calculate the area \( A \)

Calculating the integral:

\[ A = \int_0^{\frac{6}{\sqrt{13}}} \frac{1}{2}x \, dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_0^{\frac{6}{\sqrt{13}}} = \frac{1}{2} \cdot \frac{1}{2} \left( \frac{6}{\sqrt{13}} \right)^2 = \frac{1}{4} \cdot \frac{36}{13} = \frac{9}{13} \]

### Step 4: Determine the area \( A' \)

The area \( A' \) is bounded by the y-axis, the line \( y = kx \), and the ellipse. The equations are:

1. \( x = 0 \) (y-axis)
2. \( y = kx \)
3. \( \frac{x^2}{9} + \frac{y^2}{1} = 1 \)

Substituting \( y = kx \) into the ellipse equation:

\[ \frac{x^2}{9} + \frac{(kx)^2}{1} = 1 \]

This simplifies to:

\[ \frac{x^2}{9} + k^2x^2 = 1 \implies x^2\left(\frac{1}{9} + k^2\right) = 1 \implies x^2 = \frac{1}{\frac{1}{9} + k^2} \]

Thus, \( x = \sqrt{\frac{1}{\frac{1}{9} + k^2}} \).

Now substituting back to find \( y \):

\[ y = kx = k\sqrt{\frac{1}{\frac{1}{9} + k^2}} \]

### Step 5: Set up the integral for area \( A' \)

The area \( A' \) can be calculated using the integral:

\[ A' = \int_0^{\sqrt{\frac{1}{\frac{1}{9} + k^2}}} kx \, dx \]

Calculating the integral:

\[ A' = \left[ \frac{kx^2}{2} \right]_0^{\sqrt{\frac{1}{\frac{1}{9} + k^2}}} = \frac{k}{2} \left(\frac{1}{\frac{1}{9} + k^2}\right) = \frac{k}{2\left(\frac{1}{9} + k^2\right)} \]

### Step 6: Set \( A = A' \)

Setting \( A = A' \):

\[ \frac{9}{13} = \frac{k}{2\left(\frac{1}{9} + k^2\right)} \]

Cross-multiplying gives:

\[ 9 \cdot 2\left(\frac{1}{9} + k^2\right) = 13k \]

Simplifying this leads to:

\[ 18\left(\frac{1}{9} + k^2\right) = 13k \implies 2 + 18k^2 = 13k \]

Rearranging gives:

\[ 18k^2 - 13k + 2 = 0 \]

### Step 7: Solve the quadratic equation

Using the quadratic formula:

\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 18 \cdot 2}}{2 \cdot 18} \]

Calculating the discriminant:

\[ 169 - 144 = 25 \implies k = \frac{13 \pm 5}{36} \]

Thus, we have:

\[ k = \frac{18}{36} = \frac{1}{2} \quad \text{or} \quad k = \frac{8}{36} = \frac{2}{9} \]

### Step 8: Find \( 9k \)

Taking \( k = \frac{2}{9} \):

\[ 9k = 9 \cdot \frac{2}{9} = 2 \]

### Final Answer

The value of \( 9k \) such that \( A \) and \( A' \) are equal is:

\[ \boxed{2} \]