If `e_1` and `e_2` are eccentricities of `x^2 /18 + y^2 /4 = 1` and `x^2 /9 - y^2 /4 = 1` respectively and if the point `(e_1, e_2)` lies on ellipse `15x^2 + 3y^2 = k`. Then tha value of k

Options

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

Explanation

To solve the problem, we need to find the eccentricities \( e_1 \) and \( e_2 \) of the given ellipse and hyperbola, respectively, and then determine the value of \( k \) such that the point \( (e_1, e_2) \) lies on the ellipse defined by the equation \( 15x^2 + 3y^2 = k \).

### Step 1: Find the eccentricity \( e_1 \) of the ellipse \( \frac{x^2}{18} + \frac{y^2}{4} = 1 \)

The standard form of an ellipse is given by:

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

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

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

Substituting the values:

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

### Step 2: Find the eccentricity \( e_2 \) of the hyperbola \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \)

The standard form of a hyperbola is given by:

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

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

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

Substituting the values:

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

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

We need to find \( k \) such that the point \( (e_1, e_2) \) lies on the ellipse. Thus, we substitute \( e_1 \) and \( e_2 \) into the equation:

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

Calculating \( e_1^2 \) and \( e_2^2 \):

\[ \left(\frac{\sqrt{7}}{3}\right)^2 = \frac{7}{9}, \quad \left(\frac{\sqrt{13}}{3}\right)^2 = \frac{13}{9} \]

Now substituting back into the equation:

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

Calculating each term:

\[ \frac{15 \cdot 7}{9} = \frac{105}{9}, \quad \frac{3 \cdot 13}{9} = \frac{39}{9} \]

Adding these together:

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

### Final Answer

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

\[ \boxed{16} \]