The vertices a hyperbola H are ` (pm6, 0) `and its eccentricity is `sqrt5/ 2 `. Let N be the normal to H at a point in the first quadrant and parallel to the line `sqrt2 x+ y = 2sqrt 2 `. If d is the length of the line segment of N between H and the y-axis then `d^2` is equal to ______ .

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the hyperbola
The vertices of the hyperbola are given as \((\pm 6, 0)\). This means that \(a = 6\). The eccentricity \(e\) is given as \(\frac{\sqrt{5}}{2}\).

The standard form of a hyperbola centered at the origin is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
We need to find \(b\) using the relationship between \(a\), \(b\), and \(e\):
\[ e = \frac{\sqrt{a^2 + b^2}}{a} \]
Substituting the known values:
\[ \frac{\sqrt{5}}{2} = \frac{\sqrt{6^2 + b^2}}{6} \]

### Step 2: Solve for \(b^2\)
Squaring both sides:
\[ \left(\frac{\sqrt{5}}{2}\right)^2 = \left(\frac{\sqrt{36 + b^2}}{6}\right)^2 \]
This simplifies to:
\[ \frac{5}{4} = \frac{36 + b^2}{36} \]
Cross-multiplying gives:
\[ 5 \cdot 36 = 4(36 + b^2) \]
\[ 180 = 144 + 4b^2 \]
\[ 36 = 4b^2 \]
\[ b^2 = 9 \quad \Rightarrow \quad b = 3 \]

### Step 3: Write the equation of the hyperbola
Now we can write the equation of the hyperbola:
\[ \frac{x^2}{36} - \frac{y^2}{9} = 1 \]

### Step 4: Find the slope of the normal line
The line given is \(\sqrt{2}x + y = 2\sqrt{2}\). The slope \(m\) of this line is:
\[ m = -\sqrt{2} \]
The slope of the normal to the hyperbola at a point \((x_0, y_0)\) is given by:
\[ m_{\text{normal}} = \frac{b^2 x_0}{a^2 y_0} \]
Setting this equal to \(-\sqrt{2}\):
\[ -\sqrt{2} = \frac{9 x_0}{36 y_0} \quad \Rightarrow \quad -\sqrt{2} = \frac{x_0}{4 y_0} \]
Thus, we have:
\[ x_0 = -4\sqrt{2} y_0 \]

### Step 5: Substitute into the hyperbola equation
Substituting \(x_0\) into the hyperbola equation:
\[ \frac{(-4\sqrt{2} y_0)^2}{36} - \frac{y_0^2}{9} = 1 \]
This simplifies to:
\[ \frac{32 y_0^2}{36} - \frac{y_0^2}{9} = 1 \]
Finding a common denominator (36):
\[ \frac{32 y_0^2}{36} - \frac{4 y_0^2}{36} = 1 \]
\[ \frac{28 y_0^2}{36} = 1 \quad \Rightarrow \quad 28 y_0^2 = 36 \quad \Rightarrow \quad y_0^2 = \frac{36}{28} = \frac{9}{7} \]
Thus:
\[ y_0 = \frac{3}{\sqrt{7}} \]

### Step 6: Find \(x_0\)
Now substituting \(y_0\) back to find \(x_0\):
\[ x_0 = -4\sqrt{2} \cdot \frac{3}{\sqrt{7}} = -\frac{12\sqrt{2}}{\sqrt{7}} \]

### Step 7: Find the y-intercept of the normal
The equation of the normal line at point \((x_0, y_0)\) is:
\[ y - y_0 = -\sqrt{2}(x - x_0) \]
Substituting \(x_0\) and \(y_0\):
\[ y - \frac{3}{\sqrt{7}} = -\sqrt{2}\left(x + \frac{12\sqrt{2}}{\sqrt{7}}\right) \]
Setting \(x = 0\) to find the y-intercept:
\[ y - \frac{3}{\sqrt{7}} = -\sqrt{2}\left(\frac{12\sqrt{2}}{\sqrt{7}}\right) \]
\[ y = \frac{3}{\sqrt{7}} - \frac{24}{7} \]
Converting \(\frac{3}{\sqrt{7}}\) to a common denominator:
\[ y = \frac{3\sqrt{7}}{7} - \frac{24}{7} = \frac{3\sqrt{7} - 24}{7} \]

### Step 8: Calculate the length \(d\)
The length \(d\) is the distance from the point on the hyperbola to the y-axis:
\[ d = |x_0| = \frac{12\sqrt{2}}{\sqrt{7}} \]
Thus:
\[ d^2 = \left(\frac{12\sqrt{2}}{\sqrt{7}}\right)^2 = \frac{288}{7} \]

### Final Answer
Thus, \(d^2 = \frac{288}{7}\).