If straight line x cos(alpha) + y sin(alpha) = p touches the curve x2/a2 + y2/b2 = 1 , then prove that a2 cos2(alpha) + b2 sin2(alpha) = p2 .
Answers
it is given that,
xcosα + ysinα = p......(1)
x²/a² + y²/b² = 1 ........(2)
solving equations (1) and (2),
x²/a² + (pcosecα- xcotα)²/b² = 1
⇒x²b² + a²(pcosecα - xcotα)² = a²b²
⇒x²b² + a²p²cosec²α + x²a²cot²α - 2a²pcosecα.cotα x = a²b²
⇒(b² + a²cot²α)x² - 2a²pcosecα.cotα x + a²p²cosec²α - a²b² = 0
it is given that, given straight line touches the given curve so at tangent discriminanat must be equal.
i.e., D = (2a²pcosecα.cotα)² - 4(a²p²cosec²α-a²b²)(b²+a²cot²α) = 0
⇒4a⁴p²cosec²α.cot²α - 4a⁴p²cosec²α.cotα - 4a²b²p²cosec²α + 4a²b⁴ + 4a⁴b²cot²α = 0
⇒-4a²b²p²cosec²α + 4a⁴b² + 4a⁴b²cot²α = 0
⇒-a²p²cosec²α + a²b² + a²cot²α = 0
⇒a²b² + a²cot²α = a²p²cosec²α
⇒b²sin²α + a²cos²α = p² [ hence proved ]
also read similar questions: If cos-1(x/a) + cos-1(y/b) = alpha,
Prove that, x2/a2 + y2/b2 - 2xy/ab * [cos(alpha)]= sin2(alpha)
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