The curves `ax^2 + by^2 = 1` and `a'x^2 + b'y^2 = 1` intersect orthogonally if

Options

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

Answer

To determine the condition under which the curves \( ax^2 + by^2 = 1 \) and \( a'x^2 + b'y^2 = 1 \) intersect orthogonally, we need to follow these steps:

### Step 1: Differentiate the first curve
We start with the first curve:
\[ C_1: ax^2 + by^2 = 1 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(ax^2) + \frac{d}{dx}(by^2) = 0 \]
This gives us:
\[ 2ax + 2by \frac{dy}{dx} = 0 \]
Rearranging for \( \frac{dy}{dx} \):
\[ 2by \frac{dy}{dx} = -2ax \implies \frac{dy}{dx} = -\frac{ax}{by} \]
Thus, the slope \( m_1 \) of the first curve is:
\[ m_1 = -\frac{ax}{by} \]

### Step 2: Differentiate the second curve
Now we differentiate the second curve:
\[ C_2: a'x^2 + b'y^2 = 1 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(a'x^2) + \frac{d}{dx}(b'y^2) = 0 \]
This gives us:
\[ 2a'x + 2b'y \frac{dy}{dx} = 0 \]
Rearranging for \( \frac{dy}{dx} \):
\[ 2b'y \frac{dy}{dx} = -2a'x \implies \frac{dy}{dx} = -\frac{a'x}{b'y} \]
Thus, the slope \( m_2 \) of the second curve is:
\[ m_2 = -\frac{a'x}{b'y} \]

### Step 3: Condition for orthogonality
For the curves to intersect orthogonally, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the expressions for \( m_1 \) and \( m_2 \):
\[ \left(-\frac{ax}{by}\right) \cdot \left(-\frac{a'x}{b'y}\right) = -1 \]
This simplifies to:
\[ \frac{a a' x^2}{b b' y^2} = -1 \]

### Step 4: Finding the intersection points
To find the intersection points of the curves, we set:
\[ ax^2 + by^2 = a'x^2 + b'y^2 \]
Rearranging gives:
\[ (a - a')x^2 + (b - b')y^2 = 0 \]
This implies:
\[ \frac{x^2}{y^2} = \frac{b - b'}{a - a'} \]

### Step 5: Substitute into the orthogonality condition
Substituting \( \frac{x^2}{y^2} \) into the orthogonality condition:
\[ \frac{a a' \frac{b - b'}{a - a'}}{b b'} = -1 \]
Cross-multiplying and rearranging gives:
\[ a a' (b - b') = -b b' (a - a') \]

### Step 6: Final condition
Dividing both sides by \( a a' b b' \) gives:
\[ \frac{1}{b'} - \frac{1}{b} = \frac{1}{a'} - \frac{1}{a} \]
Rearranging leads to:
\[ \frac{1}{a} - \frac{1}{b} = \frac{1}{a'} - \frac{1}{b'} \]

### Final Result
Thus, the curves \( ax^2 + by^2 = 1 \) and \( a'x^2 + b'y^2 = 1 \) intersect orthogonally if:
\[ \frac{1}{a} - \frac{1}{b} = \frac{1}{a'} - \frac{1}{b'} \]