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: Understanding the Concept
When the origin is shifted to a new point $(a,b)$, the original coordinates $(x,y)$ are replaced by $(X+a, Y+b)$, where $(X,Y)$ are the new coordinates. The transformed equation given, $2X^2+XY-6Y^2+k=0$, has no first-degree terms. This means the origin has been shifted to the center of the conic section represented by the original equation. The new constant term, $k$, is the value of the original expression at the center $(a,b)$.
Step 2: Key Formula or Approach
1. Let the original equation be $f(x,y) = 2x^2+xy-6y^2-13x+9y+15=0$. 2. To find the center $(a,b)$ of the conic, we take the partial derivatives of $f(x,y)$ with respect to $x$ and $y$ and set them to zero. \[ \frac{\partial f}{\partial x} = 0 \quad \text{and} \quad \frac{\partial f}{\partial y} = 0 \] 3. Solve the resulting system of linear equations for $x=a$ and $y=b$. 4. The new constant term $k$ is given by $f(a,b)$.
Step 3: Detailed Explanation
1. Find the partial derivatives: Given $f(x,y) = 2x^2+xy-6y^2-13x+9y+15$. \[ \frac{\partial f}{\partial x} = 4x + y - 13 \] \[ \frac{\partial f}{\partial y} = x - 12y + 9 \] 2. Solve for the center (a,b): Set the partial derivatives to zero: \[ 4a + b - 13 = 0 \quad \text{(Equation 1)} \] \[ a - 12b + 9 = 0 \quad \text{(Equation 2)} \] From Equation 2, we can express $a$ in terms of $b$: \[ a = 12b - 9 \] Substitute this expression for $a$ into Equation 1: \[ 4(12b - 9) + b - 13 = 0 \] \[ 48b - 36 + b - 13 = 0 \] \[ 49b - 49 = 0 \implies b = 1 \] Now find $a$ using $a = 12b - 9$: \[ a = 12(1) - 9 = 3 \] So, the origin is shifted to the point $(a,b) = (3,1)$. 3. Calculate k: The new constant term $k$ is the value of the original function at the center $(3,1)$. \[ k = f(3,1) = 2(3)^2 + (3)(1) - 6(1)^2 - 13(3) + 9(1) + 15 \] \[ k = 2(9) + 3 - 6 - 39 + 9 + 15 \] \[ k = 18 + 3 - 6 - 39 + 9 + 15 \] \[ k = (18+3+9+15) - (6+39) = 45 - 45 = 0 \] Step 4: Final Answer
The value of the new constant term is $k=0$.