Hey plzz solve q29 Plzzz...
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hemantvats17:
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HELLO DEAR,
(x/a)cos(θ) + (y/b)sin(θ) = 1 and (x/a)sin(θ) - (y/b)cos(θ) = 1
Squaring both equations and adding them:
(x²/a²)*cos²(θ) + 2(xy/ab)cos(θ)sin(θ) + (y²/b²)*sin²(θ) + (x²/a²)sin²(θ) - 2(xy/ab)cos(θ)sin(θ) + (y²/b²)*cos²(θ) = 2
Therefore:
x²/a² + y²/b² = 2
(x/a)cos(θ) + (y/b)sin(θ) = 1 and (x/a)sin(θ) - (y/b)cos(θ) = 1
Squaring both equations and adding them:
(x²/a²)*cos²(θ) + 2(xy/ab)cos(θ)sin(θ) + (y²/b²)*sin²(θ) + (x²/a²)sin²(θ) - 2(xy/ab)cos(θ)sin(θ) + (y²/b²)*cos²(θ) = 2
Therefore:
x²/a² + y²/b² = 2
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hey buddy here is your answer
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hope it helped you
plz mark me as brainliest
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