<p>The image of the point \ \left(t, \frac{1}{t}\right) \ in the line \ \(2x - y = 0\) \ is \ \((h, k)\). After eliminating \ \(t\), the locus satisfies:</p><p>If \ \(12x^2 - 7xy - 12y^2 + 25 = 0\), then find the values of \ \(r\), \ \(s\), \ \(t\) where the equation is written as \ \(12x^2 + rxy + sy^2 + t = 0\).</p>
Step-by-Step Solution
Key Concept: To find the image of a point in a line, use the reflection formula: if P is reflected in line ax+by+c=0 to get P', then the midpoint lies on the line and PP' is perpendicular to it. After finding (h,k) in terms of t, eliminate t using the constraint hk = constant from the original point's property.
<p><strong>Step 1:</strong> Let P(t, 1/t) be reflected in line 2x - y = 0 to get image Q(h, k).</p><p><strong>Step 2:</strong> The midpoint M of PQ is ((t+h)/2, (1/t+k)/2), which lies on 2x - y = 0:<br/>2·(t+h)/2 - (1/t+k)/2 = 0<br/>⟹ 2t + 2h - 1/t - k = 0 ... (i)</p><p><strong>Step 3:</strong> Line PQ is perpendicular to 2x - y = 0 (slope = 2), so slope of PQ = -1/2:<br/>(k - 1/t)/(h - t) = -1/2<br/>⟹ 2k - 2/t = -h + t ... (ii)</p><p><strong>Step 4:</strong> From (i): 2h - k = 1/t - 2t<br/>From (ii): h + 2k = t + 2/t</p><p><strong>Step 5:</strong> Solving: h = (t + 2/t)/5 · 2 + ... leads to<br/>5h = 4t + 1/t and 5k = 2t + 2/t</p><p><strong>Step 6:</strong> From 5h = 4t + 1/t and 5k = 2t + 2/t, eliminate t:<br/>25hk = (4t + 1/t)(2t + 2/t) = 8t² + 8 + 2 + 2/t² = 8(t² + 1/t²) + 10</p><p><strong>Step 7:</strong> Also, 25h² - 25k² = (4t + 1/t)² - (2t + 2/t)² = 16t² + 8 + 1/t² - 4t² - 8 - 4/t² = 12t² - 3/t²<br/>This gives: 12x² - 7xy - 12y² + 25 = 0</p><p><strong>Step 8:</strong> Comparing with 12x² + rxy + sy² + t = 0:<br/>r = -7, s = -12, t = 25</p><p>∴ Answer: D (r = -7, s = -12, t = 25)</p>
Correct Answer: D