The straight line `ax + by = 1` makes with the curve `px^2 + 2axy + qy^2 = r`, a chord which subtends aright angle at the origin. Then

Options

A. `r(b^(2)+q^(2))=p+a`
B. `r(b^(2)+p^(2))=p+q`
C. `r(a^(2)+b^(2))=p+q`
D. `(a^(2)+p^(2))r=q+b`

Answer

To solve the problem step-by-step, we need to find the condition under which the straight line \( ax + by = 1 \) subtends a right angle at the origin with the curve \( px^2 + 2axy + qy^2 = r \).

### Step 1: Write down the equations
We have the straight line:
\[ ax + by = 1 \]
And the curve:
\[ px^2 + 2axy + qy^2 = r \]

### Step 2: Homogenize the curve equation
To homogenize the curve equation, we can substitute \( 1 \) from the line equation into the curve equation. We can express \( r \) as:
\[ r = px^2 + 2axy + qy^2 \]
We can rewrite the equation of the line as:
\[ ax + by = 1 \implies ax + by = r \cdot 1^2 \]
Thus, we have:
\[ px^2 + 2axy + qy^2 = r(ax + by)^2 \]

### Step 3: Expand the right-hand side
Expanding \( r(ax + by)^2 \):
\[ (ax + by)^2 = a^2x^2 + 2abxy + b^2y^2 \]
So, we get:
\[ r(ax + by)^2 = r(a^2x^2 + 2abxy + b^2y^2) \]
This leads to:
\[ px^2 + 2axy + qy^2 = ra^2x^2 + 2rabxy + rb^2y^2 \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ px^2 - ra^2x^2 + 2axy - 2rabxy + qy^2 - rb^2y^2 = 0 \]
This simplifies to:
\[ (p - ra^2)x^2 + (2a - 2rab)xy + (q - rb^2)y^2 = 0 \]

### Step 5: Condition for perpendicularity
For the pair of straight lines represented by the quadratic equation to be perpendicular, the condition is:
\[ \text{Coefficient of } x^2 + \text{Coefficient of } y^2 = 0 \]
Thus, we have:
\[ (p - ra^2) + (q - rb^2) = 0 \]

### Step 6: Simplify the condition
This leads to:
\[ p + q - ra^2 - rb^2 = 0 \]
Rearranging gives:
\[ p + q = ra^2 + rb^2 \]
Factoring out \( r \):
\[ p + q = r(a^2 + b^2) \]

### Final Result
Thus, the condition for the straight line \( ax + by = 1 \) to subtend a right angle at the origin with the curve \( px^2 + 2axy + qy^2 = r \) is:
\[ p + q = r(a^2 + b^2) \]