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If a hyperbola has vertices (±6, 0) and P(10, 16) lies on it, then the equation of normal at P is(1) 2x + 5y = 100 (2) 2x + 5y = 10 (3) 2x – 5y = 100 (4) 5x + 2y = 100 |
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Answer» Answer is (1) 2x + 5y = 100 Vertex is at (±6, 0) ∴ a = 6 Let the hyperbola is x2/a2 - y2/b2 = 1 Putting point P(10, 16) on the hyperbola 100/36 - 256/b2 = 1 ⇒ b2 = 144 ∴ hyperbola is x2/36 - y2/144 = 1 ∴ equation of normal is a2x/x1 + b2y/y1 = a2 + b2 ∴ putting we get 2x + 5y = 100
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