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 constant \( k \) for the ellipse described by the equation \( 15x^2 + 3y^2 = k \).

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

The equation of the ellipse is given as:

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

From this, we identify:

- \( a^2 = 18 \) (where \( a \) is the semi-major axis)
- \( b^2 = 4 \) (where \( b \) is the semi-minor axis)

The formula for the eccentricity \( e \) of an ellipse is:

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

Substituting the values:

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

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

The equation of the hyperbola is given as:

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

From this, we identify:

- \( a^2 = 9 \)
- \( b^2 = 4 \)

The formula for the eccentricity \( e \) of a hyperbola is:

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

Substituting the values:

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

### Step 3: Substitute \( (e_1, e_2) \) into the ellipse equation

We need to check if the point \( (e_1, e_2) = \left(\frac{\sqrt{7}}{3}, \frac{\sqrt{13}}{3}\right) \) satisfies the equation \( 15x^2 + 3y^2 = k \).

Substituting \( x = e_1 \) and \( y = e_2 \):

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

Calculating each term:

\[ 15\left(\frac{7}{9}\right) + 3\left(\frac{13}{9}\right) = k \]

This simplifies to:

\[ \frac{105}{9} + \frac{39}{9} = k \]

Combining the fractions:

\[ k = \frac{105 + 39}{9} = \frac{144}{9} = 16 \]

### Final Answer

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

\[ \boxed{16} \]