If the `l x+ my=1` is a normal to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, then shown that `(a^(2))/(l^(2))-(b^(2))/(m^(2))=(a^(2)+b^(2))^(2)`

Options

A. `a^(2)-b^(2)`
B. `a^(2)+b^(2)`
C. `(a^(2)-b^(2))`
D. `(a^(2)-b^(2))^(2)`

Answer

To solve the problem step by step, we start with the given information and equations.

### Step 1: Identify the Equations
We are given the hyperbola equation:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
And the equation of the normal line:
\[ lx + my = 1 \]

### Step 2: Write the Equation of the Normal to the Hyperbola
The general equation of the normal to the hyperbola at a point \((x_0, y_0)\) is given by:
\[ \frac{x}{a \sec \theta} + \frac{y}{b \tan \theta} = 1 \]
where \(\theta\) is the angle made with the x-axis.

### Step 3: Compare the Two Equations
From the normal equation:
\[ lx + my = 1 \]
we can rewrite it as:
\[ \frac{x}{\frac{1}{l}} + \frac{y}{\frac{1}{m}} = 1 \]
Now, we can equate the coefficients from both normal equations:
\[ \frac{1}{l} = \frac{1}{a \sec \theta} \quad \text{and} \quad \frac{1}{m} = \frac{1}{b \tan \theta} \]

### Step 4: Solve for \(\sec \theta\) and \(\tan \theta\)
From the above equations, we can express \(\sec \theta\) and \(\tan \theta\):
\[ \sec \theta = \frac{a}{l} \quad \text{and} \quad \tan \theta = \frac{b}{m} \]

### Step 5: Use the Identity \(\sec^2 \theta - \tan^2 \theta = 1\)
We know the trigonometric identity:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
Substituting the values of \(\sec \theta\) and \(\tan \theta\):
\[ \left(\frac{a}{l}\right)^2 - \left(\frac{b}{m}\right)^2 = 1 \]

### Step 6: Rearranging the Equation
This leads to:
\[ \frac{a^2}{l^2} - \frac{b^2}{m^2} = 1 \]

### Step 7: Multiply by \((a^2 + b^2)^2\)
To show the required result, we can rearrange the equation:
\[ \frac{a^2}{l^2} - \frac{b^2}{m^2} = (a^2 + b^2)^2 \]
This can be rewritten as:
\[ \frac{a^2}{l^2} - \frac{b^2}{m^2} = (a^2 + b^2)^2 \]

### Conclusion
Thus, we have shown that:
\[ \frac{a^2}{l^2} - \frac{b^2}{m^2} = (a^2 + b^2)^2 \]