Step 1: Find the Foci of the Ellipse

The equation of the ellipse is given by $\frac{x^2}{25} + \frac{y^2}{9} = 1$. Here, $a_e^2 = 25$ and $b_e^2 = 9$. The distance from the center to the foci is $c_e$, where $c_e^2 = a_e^2 - b_e^2$.

Calculation:

$c_e^2 = 25 - 9 = 16$
$c_e = \sqrt{16} = 4$

The foci of the ellipse are located at $(\pm c_e, 0)$, which are $(\pm 4, 0)$.

Step 2: Determine the Parameter '$a$' for the Hyperbola

The first hyperbola is given by $\frac{x^2}{a^2} - \frac{y^2}{12} = 1$. For this hyperbola, $b_h^2 = 12$. Its foci coincide with the ellipse's foci, meaning the hyperbola's foci are at $(\pm 4, 0)$.

For a hyperbola with the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the distance to the foci is $c_h$, where $c_h^2 = a^2 + b_h^2$. Since the foci coincide, $c_h = c_e = 4$.

Calculation:

$c_h^2 = a^2 + b_h^2$
$4^2 = a^2 + 12$
$16 = a^2 + 12$
$a^2 = 16 - 12 = 4$

Thus, $a = \sqrt{4} = 2$ (assuming $a>0$).

Step 3: Define the Second Hyperbola and Use the Given Point

The problem asks for the eccentricity of a hyperbola with foci at $(0, \pm \sqrt{10})$. This indicates a hyperbola with a vertical transverse axis, having the standard form $\frac{y^2}{A^2} - \frac{x^2}{B^2} = 1$. For this hyperbola, $c = \sqrt{10}$, so $c^2 = 10$. The relationship between $A$, $B$, and $c$ is $A^2 + B^2 = c^2$. Therefore, $A^2 + B^2 = 10$.

This hyperbola passes through the point $(a, a + 1)$. Using the value $a=2$ found earlier, the point is $(2, 2+1) = (2, 3)$.

Substitute the point $(2, 3)$ into the hyperbola equation $\frac{y^2}{A^2} - \frac{x^2}{B^2} = 1$:

$\frac{3^2}{A^2} - \frac{2^2}{B^2} = 1$
$\frac{9}{A^2} - \frac{4}{B^2} = 1$

Step 4: Solve for $A^2$

Substitute $B^2 = 10 - A^2$ into the equation $\frac{9}{A^2} - \frac{4}{B^2} = 1$:

$\frac{9}{A^2} - \frac{4}{10 - A^2} = 1$

Multiply both sides by $A^2(10 - A^2)$ to clear the denominators:

$9(10 - A^2) - 4A^2 = A^2(10 - A^2)$
$90 - 9A^2 - 4A^2 = 10A^2 - (A^2)^2$
$90 - 13A^2 = 10A^2 - (A^2)^2$

Rearrange into a quadratic equation for $A^2$:

$(A^2)^2 - 23A^2 + 90 = 0$

Let $X = A^2$. The equation becomes $X^2 - 23X + 90 = 0$. Factoring this quadratic gives:

$(X - 5)(X - 18) = 0$

The possible values for $A^2$ are $5$ and $18$.

• If $A^2 = 18$, then $B^2 = 10 - 18 = -8$, which is not possible for a real hyperbola.
• If $A^2 = 5$, then $B^2 = 10 - 5 = 5$. This is a valid solution.

So, $A^2 = 5$.

Step 5: Calculate the Eccentricity of the Hyperbola

The eccentricity ($e_h$) of a hyperbola with a vertical transverse axis is given by $e_h = \frac{c}{A}$.

We have $c = \sqrt{10}$ and $A = \sqrt{A^2} = \sqrt{5}$.

Calculation:

$e_h = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{\frac{10}{5}} = \sqrt{2}$

The eccentricity of the hyperbola is $\sqrt{2}$.