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Most Important Selected Qs for JEE AdvancedMathematicsHyperbola

Paragraph: The difference between the second degree curve and pair of asymptotes is constant. If second degree curve represented by a hyperbola S=0 , then the equation of its asymptotes is S+ =0 where is constant. Which will be a pair of straight lines, then we get . Then equation of asymptotes is A S+ =0 and if equation of conjugate hyperbola of S represented by S₁ , then A is the arithmetic mean of S and S₁ . Quest

Options

  1. A(2 x+3 y-3)(3 x+2 y-5)=256
  2. B(2 x+3 y-7)(3 x+2 y-8)=156
  3. C(2 x+3 y-5)(3 x+2 y-3)=252
  4. D(2 x+3 y-8)(3 x+2 y-7)=154

Correct answer

D. (2 x+3 y-8)(3 x+2 y-7)=154

Step-by-step solution

Let the asymptotes be 2 x+3 y+ =0 and 3 x+2 y+ =0 Since, asymptotes passes through (1, 2), then =-8 and =-7 Let the equation of hyperbola be (2 x+3 y-8)(3 x+2 y-7)+ =0 ...(i) It passes through (5,3) , then aligned & (10+9-8)(15+6-7)+ =0 & 11 14+ =0 & =-154 aligned Putting the value of in Eq. (i), then (2 x+3 y-8)(3 x+2 y-7)=154

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