The chords `(tx + my - 1) = 0 (t, m` being parameters) of the curve `x^2-3y^2+3xy+3x=0,` subtending a right angle at the origin, are concurrent at the point

Options

A. `(-3//2,0)`
B. `(3//2,0)`
C. `(1//2,0)`
D. none of these

Answer

To solve the problem, we need to find the point of concurrency of the chords \( tx + my - 1 = 0 \) that subtend a right angle at the origin for the given curve \( x^2 - 3y^2 + 3xy + 3x = 0 \).

### Step-by-Step Solution:

1. **Understanding the Given Chord Equation**:
The equation of the chord is given as:
\[ tx + my - 1 = 0 \]
where \( t \) and \( m \) are parameters.

2. **Homogenizing the Curve Equation**:
The equation of the curve is:
\[ x^2 - 3y^2 + 3xy + 3x = 0 \]
To find the chords that subtend a right angle at the origin, we will homogenize the equation of the chord with the curve. We can write:
\[ (x^2 - 3y^2 + 3xy + 3x)(tx + my) = 0 \]

3. **Expanding the Homogenized Equation**:
Expanding the above expression:
\[ tx(x^2 - 3y^2 + 3xy + 3x) + my(x^2 - 3y^2 + 3xy + 3x) = 0 \]
This leads to:
\[ tx^3 - 3txy^2 + 3tx^2y + 3tx^2 + myx^2 - 3my^3 + 3mxy + 3my = 0 \]

4. **Grouping Terms**:
Collecting like terms, we can express this as:
\[ x^2(t + m) + y^2(-3) + xy(3t + 3m) + 3tx = 0 \]

5. **Condition for Perpendicular Chords**:
For the lines to be perpendicular, the condition is:
\[ \text{Coefficient of } x^2 + \text{Coefficient of } y^2 = 0 \]
This gives us:
\[ 1 + 3t - 3 = 0 \]
Simplifying this, we find:
\[ 3t - 2 = 0 \implies t = \frac{2}{3} \]

6. **Substituting \( t \) Back**:
Now substituting \( t = \frac{2}{3} \) back into the chord equation:
\[ \frac{2}{3}x + my - 1 = 0 \]
Rearranging gives:
\[ 2x + 3my = 3 \]

7. **Finding the Point of Concurrency**:
Setting \( 2x - 3 = 0 \) gives:
\[ x = \frac{3}{2} \]
Substituting \( x = \frac{3}{2} \) into the rearranged equation:
\[ 2\left(\frac{3}{2}\right) + 3my = 3 \implies 3 + 3my = 3 \implies 3my = 0 \]
This implies:
\[ y = 0 \]

8. **Conclusion**:
Therefore, the point of concurrency of the chords is:
\[ \left(\frac{3}{2}, 0\right) \]