Question

If $2x^2+xy-6y^2+k=0$ is the transformed equation of $2x^2+xy-6y^2-13x+9y+15=0$ when the origin is shifted to the point $(a,b)$ by translation of axes, then k =

• A. 1
• B. 0
• C. 21
• D. 15

Hint:

For a general second-degree equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, shifting the origin to the center $(a,b)$ eliminates the first-degree terms ($x$ and $y$). The center is found by solving the system $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$. The new constant term is simply the value of the original polynomial evaluated at the center, i.e., $k = f(a,b)$.

Answer 1

The Correct Option is B

Step 1: Condition for Removing Linear Terms: When the origin is shifted to \( (a,b) \) to remove the linear terms (x and y terms) from a second-degree equation, the point \( (a,b) \) must be the center of the conic. The transformed equation will be \( Ax^2 + 2Hxy + By^2 + \Delta/ (AB-H^2) = 0 \)? Or simply, the constant term becomes \( f(a,b) \)? Not exactly. The new constant \( k \) is obtained by substituting the center \( (a,b) \) into the expression \( \frac{1}{2}(x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y}) + C \)? Actually, the simplest way is: The new constant \( k = g a + f b + c \), where g, f are half-coefficients of x, y in original, and c is original constant. Or calculate \( f(a,b) \)? No, it's \( f(a,b) \) only if we equate to new \( z \). For equation \( =0 \), the value is \( f(a,b) \). Let's verify. New equation: \( 2X^2 + XY - 6Y^2 + f(a,b) = 0 \). So \( k = f(a,b) \).
Step 2: Find the Center \( (a,b) \): Partial derivatives of \( F(x,y) = 2x^2+xy-6y^2-13x+9y+15 \): 1) \( \frac{\partial F}{\partial x} = 4x + y - 13 = 0 \) 2) \( \frac{\partial F}{\partial y} = x - 12y + 9 = 0 \) Solve the system: From (2), \( x = 12y - 9 \). Substitute into (1): \( 4(12y-9) + y - 13 = 0 \) \( 48y - 36 + y - 13 = 0 \) \( 49y = 49 \implies y = 1 \). Then \( x = 12(1) - 9 = 3 \). So, center \( (a,b) = (3,1) \).
Step 3: Calculate \( k \): \( k = F(3,1) = 2(3)^2 + (3)(1) - 6(1)^2 - 13(3) + 9(1) + 15 \) \( k = 2(9) + 3 - 6 - 39 + 9 + 15 \) \( k = 18 + 3 - 6 - 39 + 9 + 15 \) \( k = 21 - 6 - 39 + 9 + 15 \) \( k = 15 - 39 + 9 + 15 \) \( k = -24 + 24 = 0 \)