<p>If graph of \(xy = 1\) is reflected in \(y = 2x\) to give the graph \(12x^2 + rxy + sy^2 + t = 0\) then:</p>
<p>(a) \(r=1,\ s=12,\ t=25\)</p>
<p>(b) \(r=-1,\ s=12,\ t=1\)</p>
<p>(c) \(r=-7,\ s=-12,\ t=25\)</p>
<p>(d) \(r+s=-19\)</p>
Step-by-Step Solution
Key Concept: To find the reflection of a curve in a line, use the reflection transformation formulas: if point (x,y) on the original curve maps to (x',y') on the reflected curve, then (x,y) can be expressed in terms of (x',y') using the reflection geometry of the given line.
<p><strong>Step 1:</strong> For reflection in line y = 2x, use the reflection transformation. If (x,y) lies on xy=1, its reflection (x',y') satisfies the reflection formula.</p><p><strong>Step 2:</strong> The line y=2x has slope 2. For a point (x,y), its reflection (x',y') in y=2x is given by:</p><p>x' = (1-4)x + 2(2)y / (1+4) = (-3x + 4y)/5</p><p>y' = 2(2)x + (4-1)y / (1+4) = (4x + 3y)/5</p><p><strong>Step 3:</strong> Since (x,y) satisfies xy = 1, we have: x·y = 1</p><p><strong>Step 4:</strong> Express x and y in terms of x' and y' by inverting the transformation:</p><p>5x = -3x' + 4y' and 5y = 4x' + 3y'</p><p>Solve to get: x = (-3x' + 4y')/5, y = (4x' + 3y')/5</p><p><strong>Step 5:</strong> Substitute into xy = 1:</p><p>[(-3x' + 4y')/5][(4x' + 3y')/5] = 1</p><p>(-3x' + 4y')(4x' + 3y') = 25</p><p>-12x'² - 9x'y' + 16x'y' + 12y'² = 25</p><p>-12x'² + 7x'y' + 12y'² = 25</p><p><strong>Step 6:</strong> Replacing x' → x and y' → y, and rearranging to match the form 12x² + rxy + sy² + t = 0:</p><p>12x² - 7xy - 12y² + 25 = 0</p><p>Therefore: r = -7, s = -12, t = 25</p><p>∴ Answer: D</p>
Correct Answer: D