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A hyperbola with vertices at (±6, 0) passes through the point P(10, 16). Find the equation of the normal to the hyperbola at P.
- 2x + 5y = 100
- x + 2y = 42
- 3x + 4y = 94
- x + 3y = 58
Correct answer: 2x + 5y = 100
Solution
a² = 36; from P: 100/36 - 256/b² = 1 gives b² = 144. Slope of tangent at P from x x1/a² - y y1/b² = 1 leads to a normal 2x + 5y = 100.
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