If `ax^(2)+by^(2)=1` cut `a'x^(2)+b'y^(2)=1` orthogonally, then

Options

A. `(1)/(a)-(1)/(a')=(1)/(b)-(1)/(b')`
B. `(1)/(a)+(1)/(a')=(1)/(b)+(1)/(b')`
C. `(1)/(a)+(1)/(b)=(1)/(a')+(1)/(b')`
D. None of these

Answer

To solve the problem where the curves \( ax^2 + by^2 = 1 \) and \( a'x^2 + b'y^2 = 1 \) intersect orthogonally, we need to find the relationship between the coefficients \( a, a', b, \) and \( b' \). Here’s a step-by-step solution:

### Step 1: Differentiate the equations
We start by differentiating both equations with respect to \( x \).

1. For the first equation \( ax^2 + by^2 = 1 \):
\[ \frac{d}{dx}(ax^2) + \frac{d}{dx}(by^2) = 0 \]
This gives:
\[ 2ax + 2by \frac{dy}{dx} = 0 \]
Rearranging, we find:
\[ \frac{dy}{dx} = -\frac{ax}{by} \]

2. For the second equation \( a'x^2 + b'y^2 = 1 \):
\[ \frac{d}{dx}(a'x^2) + \frac{d}{dx}(b'y^2) = 0 \]
This gives:
\[ 2a'x + 2b'y \frac{dy}{dx} = 0 \]
Rearranging, we find:
\[ \frac{dy}{dx} = -\frac{a'x}{b'y} \]

### Step 2: Set up the orthogonality condition
Since the curves intersect orthogonally, the product of their slopes must equal \(-1\):
\[ \left(-\frac{ax}{by}\right) \left(-\frac{a'x}{b'y}\right) = -1 \]
This simplifies to:
\[ \frac{a a' x^2}{b b' y^2} = -1 \]

### Step 3: Rearranging the equation
From the equation above, we can rearrange it to obtain:
\[ a a' x^2 = -b b' y^2 \]

### Step 4: Subtract the equations
Next, we can subtract the two original equations:
\[ (ax^2 + by^2) - (a'x^2 + b'y^2) = 0 \]
This simplifies to:
\[ (a - a')x^2 + (b - b')y^2 = 0 \]

### Step 5: Compare coefficients
From the equations \( a a' x^2 = -b b' y^2 \) and \( (a - a')x^2 + (b - b')y^2 = 0 \), we can compare coefficients.

Dividing the two equations gives:
\[ \frac{a - a'}{a a'} = \frac{b - b'}{-b b'} \]

### Step 6: Final relationship
Cross-multiplying leads to:
\[ (a - a')b b' = -(b - b')a a' \]
Rearranging gives us:
\[ \frac{1}{a} - \frac{1}{a'} = \frac{1}{b} - \frac{1}{b'} \]

### Conclusion
Thus, the relationship between the coefficients is:
\[ \frac{1}{a} - \frac{1}{a'} = \frac{1}{b} - \frac{1}{b'} \]