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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. A10 sq unit
  2. B20 sq unit
  3. C30 sq unit
  4. D40 sq unit

Correct answer

B. 20 sq unit

Step-by-step solution

aligned & 16 x^2-25 y^2=400 & x^2 5^2 - y^2 4^2 =1 ...(i) aligned Let P (5 , 4 ) be any point on the hyperbola (i) Equation of tangent at P is x 5 - y 4 =1 ...(ii) And asymptotes of Eq. (i) are y= 4 5 x ...(iii) Solving Eqs. (ii) and (iii), then x 5 x 5 =1 aligned or x & = 5 ( ) & = 5( + )( - ) ( ) aligned then we get A [5( + ), 4( + )] and B [(5( - ),-4( - )] Area of ABC = 1 2 |-20-20|=20 sq unit

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