JEE Main 2024 Mathematics question
If x² – y² + 2hxy + 2gx + 2fy + c = 0 is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals
Answer: (C)
Solution
Equal distances give (x + 2y + 7)² = (2x − y + 8)² (both have the same denominator, 5). Expanding, −3x² + 3y² + 8xy − 18x + 44y − 15 = 0, i.e. x² − y² −
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