If the straight lines joining the origin and the points of intersection ofthe curve `5x^2 + 1 2xy-6y^2 + 4x-2y+ 3=0 and x + ky-1=0` are equally inclined to the co-ordinate axes then the value of k

Options

A. is equal to 1
B. is equal to -1
C. is equal to 2
D. does not exist in th set of real numbers.

Answer

To find the value of \( k \) such that the straight lines joining the origin and the points of intersection of the given curve and line are equally inclined to the coordinate axes, we can follow these steps:

### Step 1: Write down the equations
We have the curve given by:
\[ 5x^2 + 12xy - 6y^2 + 4x - 2y + 3 = 0 \quad \text{(Equation 1)} \]
and the line given by:
\[ x + ky - 1 = 0 \quad \text{(Equation 2)} \]

### Step 2: Substitute the line equation into the curve equation
From Equation 2, we can express \( x \) in terms of \( y \):
\[ x = 1 - ky \]
Now, substitute \( x = 1 - ky \) into Equation 1:
\[ 5(1 - ky)^2 + 12(1 - ky)y - 6y^2 + 4(1 - ky) - 2y + 3 = 0 \]

### Step 3: Expand and simplify the equation
Expanding the terms:
\[ 5(1 - 2ky + k^2y^2) + 12y - 12ky - 6y^2 + 4 - 4ky - 2y + 3 = 0 \]
This simplifies to:
\[ 5 - 10ky + 5k^2y^2 + 12y - 12ky - 6y^2 + 4 - 4ky - 2y + 3 = 0 \]
Combining like terms:
\[ (5k^2 - 6)y^2 + (-10k - 12k - 4k + 12)y + (5 + 4 + 3) = 0 \]
This simplifies to:
\[ (5k^2 - 6)y^2 + (-26k + 12)y + 12 = 0 \]

### Step 4: Identify the condition for equal inclination
For the lines to be equally inclined to the axes, the coefficient of \( xy \) must be zero. The coefficient of \( xy \) in our equation is \( -26k + 12 \). Setting this equal to zero gives:
\[ -26k + 12 = 0 \]

### Step 5: Solve for \( k \)
Rearranging the equation:
\[ 26k = 12 \implies k = \frac{12}{26} = \frac{6}{13} \]

### Step 6: Conclusion
Thus, the value of \( k \) is:
\[ \boxed{\frac{6}{13}} \]