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AP EAMCET201922 Apr 2019Morning ShiftMathematicsParabolaActual

The length of the latusrectum of the parabola (20 (x^2+y^2-6 x-2 y+10 )=(4 x-2 y-5)^2 ), is

Options

  1. A( 5 2 )
  2. B(2 5 )
  3. C( 5 )
  4. D(4 5 )

Correct answer

C. ( 5 )

Step-by-step solution

Given, equation of parabola is ( aligned 20 (x^2+y^2-6 x-2 y+10 ) & =(4 x-2 y-5)^2 (x^2+y^2-6 x-2 y+10 ) & = ( 4 x-2 y-5 20 )^2 (x-3)^2+(y-1)^2 & = ( 4 x-2 y-5 20 )^2 (i) aligned ) In Eq. (i), focus is ((3,1) ) and equation of directrix is (4 x-2 y-5=0 ). So, distance from focus to directrix is ( |12-2-5| 20 ) (= 5 2 =2 a ) Now, length of latursrectum (=4 a ) (=2(2 a)=2 5 2 = 5 )

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