Equation of normal to `9x^(2)-25y^(2)=225` at `theta=pi//4` is

Options

A. `5x+3 sqrt2y=34sqrt2`
B. `5x+sqrt2y=34sqrt2`
C. `5x+sqrt3y=34sqrt2`
D. `5x-3sqrt2y=34sqrt2`

Answer

To find the equation of the normal to the hyperbola given by \( 9x^2 - 25y^2 = 225 \) at \( \theta = \frac{\pi}{4} \), we can follow these steps:

### Step 1: Convert the hyperbola equation to standard form
We start with the equation of the hyperbola:
\[ 9x^2 - 25y^2 = 225 \]
Dividing the entire equation by 225 gives:
\[ \frac{9x^2}{225} - \frac{25y^2}{225} = 1 \]
This simplifies to:
\[ \frac{x^2}{25} - \frac{y^2}{9} = 1 \]

### Step 2: Identify \( a \) and \( b \)
From the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we can identify:
\[ a^2 = 25 \quad \Rightarrow \quad a = 5 \]
\[ b^2 = 9 \quad \Rightarrow \quad b = 3 \]

### Step 3: Find the coordinates of the point on the hyperbola
Using \( \theta = \frac{\pi}{4} \), we find the coordinates \( (x, y) \) using the parametric equations:
\[ x = a \sec \theta = 5 \sec \left(\frac{\pi}{4}\right) = 5 \sqrt{2} \]
\[ y = b \tan \theta = 3 \tan \left(\frac{\pi}{4}\right) = 3 \]

### Step 4: Write the equation of the normal
The equation of the normal to the hyperbola at the point \( (a \sec \theta, b \tan \theta) \) is given by:
\[ \frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2 + b^2 \]
Substituting the values:
\[ \frac{5x}{\sec \left(\frac{\pi}{4}\right)} + \frac{3y}{\tan \left(\frac{\pi}{4}\right)} = 5^2 + 3^2 \]
This simplifies to:
\[ \frac{5x}{\sqrt{2}} + 3y = 25 + 9 \]
\[ \frac{5x}{\sqrt{2}} + 3y = 34 \]

### Step 5: Clear the fraction
Multiplying through by \( \sqrt{2} \) to eliminate the fraction:
\[ 5x + 3\sqrt{2}y = 34\sqrt{2} \]

### Final Equation
Thus, the equation of the normal to the hyperbola at \( \theta = \frac{\pi}{4} \) is:
\[ 5x + 3\sqrt{2}y = 34\sqrt{2} \]

---