Question

After the coordinate axes are rotated through an angle \(\frac{\pi}{4}\) in the anticlockwise direction without shifting the origin, if the equation \(x^2 + y^2 - 2x - 4y - 20 = 0\) transforms to \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) in the new coordinate system, then
\[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \]

• A. \(-20\)
• B. \(-25\)
• C. \(-30\)
• D. \(-35\)

Hint:

For a general second-degree equation \(Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0\), the determinant \(\Delta = \begin{vmatrix} A & H & G
H & B & F
G & F & C \end{vmatrix}\) is an invariant. This means its value remains unchanged under rotation and translation of the coordinate axes. This property can significantly simplify problems asking for the value of this determinant after such transformations, as you can calculate it directly from the coefficients of the original equation.

Answer 1

The Correct Option is B

Method 1: Using Coordinate Transformation Step 1: Write down the transformation equations for rotation of axes.
When the coordinate axes are rotated through an angle \(\theta\) in the anti-clockwise direction without shifting the origin, the relationship between the old coordinates \((x, y)\) and the new coordinates \((x', y')\) is given by: \[ x = x' \cos\theta - y' \sin\theta \] \[ y = x' \sin\theta + y' \cos\theta \] Given \(\theta = \frac{\pi}{4}\), we have \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\) and \(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\). Substituting these values: \[ x = \frac{x'}{\sqrt{2}} - \frac{y'}{\sqrt{2}} = \frac{x' - y'}{\sqrt{2}} \] \[ y = \frac{x'}{\sqrt{2}} + \frac{y'}{\sqrt{2}} = \frac{x' + y'}{\sqrt{2}} \] Step 2: Substitute the transformation equations into the original equation.
The original equation is \(x^2 + y^2 - 2x - 4y - 20 = 0\).
Substitute the expressions for \(x\) and \(y\):
\[ \left(\frac{x' - y'}{\sqrt{2}}\right)^2 + \left(\frac{x' + y'}{\sqrt{2}}\right)^2 - 2\left(\frac{x' - y'}{\sqrt{2}}\right) - 4\left(\frac{x' + y'}{\sqrt{2}}\right) - 20 = 0 \] Expand the squares: \[ \frac{(x')^2 - 2x'y' + (y')^2}{2} + \frac{(x')^2 + 2x'y' + (y')^2}{2} - \frac{2(x' - y')}{\sqrt{2}} - \frac{4(x' + y')}{\sqrt{2}} - 20 = 0 \] Combine the terms with common denominators: \[ \frac{2(x')^2 + 2(y')^2}{2} - \sqrt{2}(x' - y') - 2\sqrt{2}(x' + y') - 20 = 0 \] \[ (x')^2 + (y')^2 - \sqrt{2}x' + \sqrt{2}y' - 2\sqrt{2}x' - 2\sqrt{2}y' - 20 = 0 \] Combine like terms: \[ (x')^2 + (y')^2 + (-\sqrt{2} - 2\sqrt{2})x' + (\sqrt{2} - 2\sqrt{2})y' - 20 = 0 \] \[ (x')^2 + (y')^2 - 3\sqrt{2}x' - \sqrt{2}y' - 20 = 0. \] Step 3: Identify the coefficients \(a, b, h, g, f, c\) in the transformed equation.
The transformed equation is of the form \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\).
Comparing \((x')^2 + (y')^2 - 3\sqrt{2}x' - \sqrt{2}y' - 20 = 0\) with the general form (using \(x'\) and \(y'\)):
\(a = 1\) (coefficient of \((x')^2\))
\(b = 1\) (coefficient of \((y')^2\))
\(2h = 0 \implies h = 0\) (coefficient of \(x'y'\))
\(2g = -3\sqrt{2} \implies g = -\frac{3\sqrt{2}}{2}\) (coefficient of \(x'\))
\(2f = -\sqrt{2} \implies f = -\frac{\sqrt{2}}{2}\) (coefficient of \(y'\))
\(c = -20\) (constant term) Step 4: Calculate the determinant \(\begin{vmatrix} a & h & g
h & b & f
g & f & c \end{vmatrix}\).
Substitute the identified coefficients into the determinant: \[ \begin{vmatrix} 1 & 0 & -\frac{3\sqrt{2}}{2} \\ 0 & 1 & -\frac{\sqrt{2}}{2} \\ -\frac{3\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & -20 \end{vmatrix} \] Expand the determinant along the first row: \[ = 1 \left( (1)(-20) - \left(-\frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{2}}{2}\right) \right) - 0 \left( \text{minor} \right) + \left(-\frac{3\sqrt{2}}{2}\right) \left( (0)\left(-\frac{\sqrt{2}}{2}\right) - (1)\left(-\frac{3\sqrt{2}}{2}\right) \right) \] \[ = 1 \left( -20 - \frac{2}{4} \right) + \left(-\frac{3\sqrt{2}}{2}\right) \left( 0 + \frac{3\sqrt{2}}{2} \right) \] \[ = \left( -20 - \frac{1}{2} \right) - \left(\frac{3\sqrt{2}}{2}\right) \left(\frac{3\sqrt{2}}{2}\right) \] \[ = -\frac{41}{2} - \frac{9 \times 2}{4} \] \[ = -\frac{41}{2} - \frac{18}{4} \] \[ = -\frac{41}{2} - \frac{9}{2} \] \[ = -\frac{50}{2} = -25. \] Method 2: Using Invariants (Shorter Method) The value of the determinant \(\Delta = \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix}\) from the general second-degree equation \(Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0\) is an invariant under rotation and translation of axes. This means its value does not change after coordinate transformation.
The given original equation is \(x^2 + y^2 - 2x - 4y - 20 = 0\).
Comparing this to \(Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0\):
\(A = 1\) (coefficient of \(x^2\))
\(B = 1\) (coefficient of \(y^2\))
\(2H = 0 \implies H = 0\) (coefficient of \(xy\))
\(2G = -2 \implies G = -1\) (coefficient of \(x\))
\(2F = -4 \implies F = -2\) (coefficient of \(y\))
\(C = -20\) (constant term) Now, calculate the determinant using these original coefficients: \[ \Delta = \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} = \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ -1 & -2 & -20 \end{vmatrix} \] Expand the determinant along the first row: \[ = 1 \left( (1)(-20) - (-2)(-2) \right) - 0 \left( \text{minor} \right) + (-1) \left( (0)(-2) - (1)(-1) \right) \] \[ = 1 \left( -20 - 4 \right) - 0 - 1 \left( 0 + 1 \right) \] \[ = -24 - 1 \] \[ = -25. \] Since the determinant is an invariant, the value of \(\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}\) in the new coordinate system is also -25.