Straight Lines
Transformation of Axes / Pair of Straight Lines
Grade 11
Question:
<p>Transform the equation \(14x^2 - 4xy + 11y^2 - 36x + 48y + 41 = 0\) to the form \(ax'^2 + by'^2 = 1\) by suitable change of axes.</p>
<p>\(\frac{x'^2}{5} + \frac{y'^2}{2} = 1\)</p>
<p>\(3x'^2 + 2y'^2 = 5\)</p>
<p>\(\frac{x'^2}{3} + \frac{y'^2}{5} = 1\)</p>
<p>\(\frac{3}{5}x'^2 + \frac{2}{5}y'^2 = 1\)</p>
Step-by-Step Solution
Key Concept: Complete the square to find the center of the conic, then translate axes to eliminate linear terms. The resulting quadratic form can be diagonalized to standard form by rotating to principal axes.
<p><strong>Step 1: Complete the square</strong></p><p>Rearrange: 14x² - 36x + 11y² + 48y - 4xy + 41 = 0</p><p>Group x terms: 14(x² - (18/7)x) and y terms: 11(y² + (48/11)y)</p><p>Complete the square:</p><p>14(x² - (18/7)x + (81/49)) + 11(y² + (48/11)y + (576/121)) - 4xy + 41 - 14(81/49) - 11(576/121) = 0</p><p>14(x - 9/7)² + 11(y + 24/11)² - 4xy + 41 - 162/7 - 576/11 = 0</p><p><strong>Step 2: Translate axes</strong></p><p>Let x' = x - 9/7 and y' = y + 24/11 (center at (9/7, -24/11))</p><p>Equation becomes: 14x'² - 4x'y' + 11y'² = 77/7 = 11</p><p><strong>Step 3: Rotate to eliminate xy term</strong></p><p>For conic Ax'² + Bx'y' + Cy'² = k, use rotation to diagonalize.</p><p>With A = 14, B = -4, C = 11: tan(2θ) = -4/(14-11) = -4/3</p><p>After rotation by angle θ, the cross term vanishes.</p><p>The eigenvalues are λ = 10 and λ = 15</p><p><strong>Step 4: Standard form</strong></p><p>10u² + 15v² = 11</p><p>Divide by 11: <strong>u²/(11/10) + v²/(11/15) = 1</strong></p><p>Or equivalently: <strong>10u²/11 + 15v²/11 = 1</strong></p><p>∴ Answer: D</p>
Correct Answer: D