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Most Important Selected Qs for JEE AdvancedMathematicsParabola

The parabola x=y^2+a y+b intersect the parabola x^2=y at (1,1) at right angle. Which of the following is/are correct?

Options

  1. Aa=4, b=-4
  2. Ba=2, b=-2
  3. CEquation of the director circle for the parabola x=y^2+a y+b is 4 x+1=0 .
  4. DArea enclosed by the parabola x=y^2+a y+b and its latus rectum is 1 6

Correct answer

D. Area enclosed by the parabola x=y^2+a y+b and its latus rectum is 1 6

Step-by-step solution

We have x=y^2+a y+b ....(1) and x^2=y ....(2) As (1) passing through (1,1) aligned & 1=1+a+b & a+b=0 aligned Now, 1=2 y dy dx + a dy dx . d y d x ]_ (x₁, y₁ ) = 1 2+a and for .y=x^2, d y d x =2 x d y d x ]_ (1,1) =2 aligned & 2 2+a =-1 & -2=2+a & a=-4 and b=4 aligned Hence parabola is y ^2-4 y +4= x ( y -2)^2= x Let y -2= Y Y ^2= X Hence Required area =2 ₀^ 1 4 x d x=(2) ( 2 3 ) (x^ 3 2 )₀^ 1 4 = 1 6 sq. units Ans. (D) Equation of directrix is x= -1 4 4 x+1=0

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