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Let \( A(\sec \theta, 2 \tan \theta) \) and \( B(\sec \phi, 2 \tan \phi) \), where \( \theta+\phi=\pi / 2 \), be two points on the hyperbola \( 2 x^{2}-y^{2}=2 \). If \( (\alpha, \beta) \) is the point of the intersection of the normals to the hyperbola at \( A \) and \( B \), then \( (2 \beta)^{2} \) is equal to |
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Answer» ans. 36 since, 2x - y = 2 x²/a²-y²/b² = 1 equation of normal, a²x/x1 + b²y/y₁ = a² + b² equation of normal at point A (secθ, 2tanθ) a² = 1, b² = 2 x1 = secθ, y₁ = 2tanθ 1 × X/secθ + 2 × y/2tanθ = 1 + 2 x/secθ + y/tanθ = 3 x/secθ + ycosθ/sinθ = 3 x/secθ + ycosθ × coseceθ = 3 x/secθ + ycosecθ/secθ = 3 1/secθ(x + ycosecθ) = 3 x + ycosecθ = 3secθ _________(i) Similarly, equation of normal at point B(secΦ, 2tanΦ) x + ycosecΦ = 3secΦ θ + Φ = π/2 Φ = π/2 - θ x + усosес(π/2 - θ) = 3sес(π/2 - θ) x + ysecθ = 3cosecθ _________(ii) (i) - (ii) x + ycosecθ = 3secθ - (x + ysecθ = 3cosecθ) y(cosecθ - secθ) = -3(cosecθ - secθ) y = -3 x²/a² - y²/b² = 1 x²/1 - (-3)²/2 = 1 x2 - 9/2 = 1 x² = 1 + 9/2 x = √/11/2 x,y = α,β = (√11/2, -3) β = -3 (2β)2 = {2 × (-3)}2 = 36
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