If `e_(1) and e_(2)` are the eccentricities of the ellipse `(x^(2))/(18)+(y^(2))/(4)=1` and the hyperbola `(x^(2))/(9)-(y^(2))/(4)=1` respectively and `(e_(1), e_(2))` is a point on the ellipse `15x^(2)+3y^(2)=k`, then the value of k is equal to

Options

A. 16
B. 17
C. 15
D. 14

Answer

To solve the problem, we need to find the eccentricities of the given ellipse and hyperbola, and then use these values to find the value of \( k \) for the ellipse \( 15x^2 + 3y^2 = k \).

### Step 1: Find the eccentricity \( e_1 \) of the ellipse

The equation of the ellipse is given by:

\[ \frac{x^2}{18} + \frac{y^2}{4} = 1 \]

Here, \( a^2 = 18 \) and \( b^2 = 4 \). The eccentricity \( e_1 \) of an ellipse is given by:

\[ e_1 = \sqrt{1 - \frac{b^2}{a^2}} \]

Calculating \( e_1 \):

\[ e_1 = \sqrt{1 - \frac{4}{18}} = \sqrt{1 - \frac{2}{9}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3} \]

### Step 2: Find the eccentricity \( e_2 \) of the hyperbola

The equation of the hyperbola is given by:

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

Here, \( a^2 = 9 \) and \( b^2 = 4 \). The eccentricity \( e_2 \) of a hyperbola is given by:

\[ e_2 = \sqrt{1 + \frac{b^2}{a^2}} \]

Calculating \( e_2 \):

\[ e_2 = \sqrt{1 + \frac{4}{9}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \]

### Step 3: Substitute \( e_1 \) and \( e_2 \) into the ellipse equation \( 15x^2 + 3y^2 = k \)

We have found:

\[ e_1 = \frac{\sqrt{7}}{3}, \quad e_2 = \frac{\sqrt{13}}{3} \]

Now, we substitute these values into the equation of the ellipse:

\[ 15\left(\frac{\sqrt{7}}{3}\right)^2 + 3\left(\frac{\sqrt{13}}{3}\right)^2 = k \]

Calculating \( k \):

\[ 15 \cdot \frac{7}{9} + 3 \cdot \frac{13}{9} = k \]

\[ = \frac{15 \cdot 7 + 3 \cdot 13}{9} = \frac{105 + 39}{9} = \frac{144}{9} = 16 \]

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{16} \]