Then angle of intersection of the normal at the point `(-5/(sqrt(2)),3/(sqrt(2)))` of the curves `x^2-y^2=8` and `9x^2+25 y^2=225` is 0 (b) `pi/2` (c) `pi/3` (d) `pi/4`

Options

A. 0
B. `(pi)/(2)`
C. `(pi)/(3)`
D. `(pi)/(4)`

Answer

To find the angle of intersection of the normals at the point \((-5/\sqrt{2}, 3/\sqrt{2})\) of the curves \(x^2 - y^2 = 8\) and \(9x^2 + 25y^2 = 225\), we will follow these steps:

### Step 1: Find the slope of the normal to the first curve
The first curve is given by:
\[ x^2 - y^2 = 8 \]
Differentiating implicitly with respect to \(x\):
\[ 2x - 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{y} \]
The slope of the normal, \(m_1\), is the negative reciprocal of \(\frac{dy}{dx}\):
\[ m_1 = -\frac{y}{x} \]

### Step 2: Calculate \(m_1\) at the given point
Substituting the point \((-5/\sqrt{2}, 3/\sqrt{2})\) into the expression for \(m_1\):
\[ m_1 = -\frac{3/\sqrt{2}}{-5/\sqrt{2}} = \frac{3}{5} \]

### Step 3: Find the slope of the normal to the second curve
The second curve is given by:
\[ 9x^2 + 25y^2 = 225 \]
Differentiating implicitly with respect to \(x\):
\[ 18x + 50y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{9x}{25y} \]
The slope of the normal, \(m_2\), is the negative reciprocal of \(\frac{dy}{dx}\):
\[ m_2 = \frac{25y}{9x} \]

### Step 4: Calculate \(m_2\) at the given point
Substituting the point \((-5/\sqrt{2}, 3/\sqrt{2})\) into the expression for \(m_2\):
\[ m_2 = \frac{25 \cdot (3/\sqrt{2})}{9 \cdot (-5/\sqrt{2})} = \frac{75/\sqrt{2}}{-45/\sqrt{2}} = -\frac{5}{3} \]

### Step 5: Check if the normals are orthogonal
To check if the normals are orthogonal, we calculate the product of the slopes:
\[ m_1 \cdot m_2 = \left(\frac{3}{5}\right) \cdot \left(-\frac{5}{3}\right) = -1 \]
Since the product is \(-1\), the normals are orthogonal.

### Step 6: Find the angle of intersection
The angle between two orthogonal lines is \(\frac{\pi}{2}\).

### Final Answer
Thus, the angle of intersection of the normals at the given point is:
\[ \frac{\pi}{2} \]