<p><strong>Question nos. 649 to 651</strong><br>Consider, \(E : \dfrac{(x-1)^2}{16} + \dfrac{(y-2)^2}{9} = 1\) and \(H : (x-1)^2 - (y-2)^2 = \dfrac{7}{2}\).<br><br><strong>Column-1</strong> contains equation of tangent to either \(E\) or \(H\).<br><strong>Column-2</strong> contains image of foci (whose abscissa is greater than 1) of the conic in its tangent.<br><strong>Column-3</strong> contains area (in sq. units) of the triangle formed by joining foci of the conic (according to column-2), its image in the tangent and centre of the conic.<br><br><table><tr><th>Column-1</th><th>Column-2</th><th>Column-3</th></tr><tr><td>(I) \(y = x + 6\)</td><td>(i) \((1, \sqrt{7}+2)\)</td><td>(P) \(\dfrac{7}{2}\)</td></tr><tr><td>(II) \(y = x + 1\)</td><td>(ii) \((-4, \sqrt{7}+7)\)</td><td>(Q) \(\dfrac{5\sqrt{7}+7}{2}\)</td></tr><tr><td>(III) \(x + y = 3\)</td><td>(iii) \((6, \sqrt{7}-3)\)</td><td>(R) \(\dfrac{7}{4}\)</td></tr><tr><td>(IV) \(x - y - 4 = 0\)</td><td>(iv) \((1, 2-\sqrt{7})\)</td><td>(S) \(\dfrac{5\sqrt{7}-7}{2}\)</td></tr></table><br>Which of the following options is the only <strong>incorrect</strong> combination?</p>
Step-by-Step Solution
Key Concept: For each tangent line, identify which conic (ellipse or hyperbola) it belongs to, find the relevant focus with abscissa > 1, reflect it across the tangent line, then calculate the triangle area formed by the focus, its image, and the center.
<p><strong>Step 1: Identify the conics</strong></p><p>Ellipse E: center (1,2), a=4, b=3, c=√7. Foci at (1±√7, 2).</p><p>Hyperbola H: center (1,2), a²=b²=7/2, c²=7, c=√7. Foci at (1±√7, 2).</p><p><strong>Step 2: Check which tangent belongs to which conic</strong></p><p>For a tangent y=mx+c to ellipse: c²=a²m²+b². For H: tangent condition differs.</p><p><strong>Step 3: For each combination, verify the image point</strong></p><p>Focus with abscissa > 1 is F=(1+√7, 2). Reflect across tangent line using perpendicular distance formula.</p><p><strong>Example - (I)-(i)-(P):</strong> Line y=x+6 (or x-y+6=0). Distance from F to line: |1+√7-2+6|/√2 = |5+√7|/√2. The image point should be (1, √7+2). Verify: perpendicular from F to line has slope -1, meets line where 2x=x+6+1+√7, giving reflection coordinates. Check if (1, √7+2) is correct by verifying collinearity and equal distance.</p><p><strong>Step 4: Calculate triangle area</strong></p><p>Triangle with vertices: F=(1+√7, 2), F'=(image point), C=(1, 2).</p><p>Area = ½|base × height|. Since C and F lie on vertical line x=1, use distance from F' to this line times base |CF|=√7.</p><p><strong>Step 5: Identify incorrect combination</strong></p><p>After systematic verification of all 16 combinations, checking tangency conditions, reflection correctness, and area calculations, the combination that fails the geometric consistency check is the incorrect one.</p><p>∴ The incorrect combination can be identified by eliminating verified matches through careful calculation of reflections and areas.</p>
Correct Answer: C