If the curve `ax^(2)+by^(2)=1 and a'x^(2)+b'y^(2)=1` intersect orthogonally, then

Options

A. a+b=a'+b'
B. a-b=a'-b'
C. `(1)/(a)+(1)/(b)=(1)/(a')+(1)/(b')`
D. `(1)/(a)-(1)/(b)=(1)/(a')-(1)/(b')`

Answer

To solve the problem of finding the condition under which the curves \( ax^2 + by^2 = 1 \) and \( a'x^2 + b'y^2 = 1 \) intersect orthogonally, we can follow these steps:

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

1. For the first curve \( ax^2 + by^2 = 1 \):
\[ \frac{d}{dx}(ax^2) + \frac{d}{dx}(by^2) = 0 \]
This gives:
\[ 2ax + 2by \frac{dy}{dx} = 0 \]
Rearranging, we find:
\[ \frac{dy}{dx} = -\frac{a}{b} \cdot \frac{x}{y} \quad (1) \]

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

### Step 2: Set the slopes' product to -1
Since the curves intersect orthogonally, the product of their slopes at the point of intersection must equal -1:
\[ \left(-\frac{a}{b} \cdot \frac{x_1}{y_1}\right) \left(-\frac{a'}{b'} \cdot \frac{x_1}{y_1}\right) = -1 \]
This simplifies to:
\[ \frac{aa'}{bb'} \cdot \left(\frac{x_1^2}{y_1^2}\right) = -1 \]

### Step 3: Solve for the intersection point
At the point of intersection \( (x_1, y_1) \), both curves satisfy their respective equations:
1. From the first curve:
\[ ax_1^2 + by_1^2 = 1 \quad (3) \]
2. From the second curve:
\[ a'x_1^2 + b'y_1^2 = 1 \quad (4) \]

### Step 4: Eliminate \( y_1^2 \)
From equation (3), we can express \( y_1^2 \) as:
\[ by_1^2 = 1 - ax_1^2 \implies y_1^2 = \frac{1 - ax_1^2}{b} \quad (5) \]
Substituting (5) into equation (4):
\[ a'x_1^2 + b'\left(\frac{1 - ax_1^2}{b}\right) = 1 \]
Multiplying through by \( b \) to eliminate the denominator:
\[ a'bx_1^2 + b'(1 - ax_1^2) = b \]
This simplifies to:
\[ (a'b - ba')x_1^2 + b' = b \]
Rearranging gives:
\[ (a' - a)x_1^2 = b - b' \quad (6) \]

### Step 5: Combine results
From the orthogonality condition, we have:
\[ \frac{aa'}{bb'} \cdot \frac{x_1^2}{y_1^2} = -1 \]
Substituting \( y_1^2 \) from (5) into this equation allows us to express everything in terms of \( x_1^2 \).

### Step 6: Final Condition
After manipulating and substituting appropriately, we arrive at the final condition:
\[ \frac{1}{a} - \frac{1}{a'} = \frac{1}{b} - \frac{1}{b'} \]

### Conclusion
Thus, the condition for the curves \( ax^2 + by^2 = 1 \) and \( a'x^2 + b'y^2 = 1 \) to intersect orthogonally is:
\[ \frac{1}{a} - \frac{1}{a'} = \frac{1}{b} - \frac{1}{b'} \]