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The orthogonal trejections of the family of parabola’s y2 = 4ax + 4a2 is the family
  • a)
    of circles
  • b)
    of ellipse
  • c)
    itself
  • d)
    of hyper-bola’s
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The orthogonal trejections of the family of parabola’s y2 = 4ax ...
we have y2 = 4ax + 4a   ...(1)
 put it in (1), we have
⇒ y = 2 xy’ + y(y')2    ...(2) ;  DE o f given family
now replace  in (2) , to get the DE. of orthogonal trajectories,
which is the same as the differential equation (2) of the given system (1). Hence the system of parabolas (1) is self orthogonal i.e. each member of the given family of parabola's intersects its own member orthogonally.
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Most Upvoted Answer
The orthogonal trejections of the family of parabola’s y2 = 4ax ...
we have y2 = 4ax + 4a   ...(1)
 put it in (1), we have
⇒ y = 2 xy’ + y(y')2    ...(2) ;  DE o f given family
now replace  in (2) , to get the DE. of orthogonal trajectories,
which is the same as the differential equation (2) of the given system (1). Hence the system of parabolas (1) is self orthogonal i.e. each member of the given family of parabola's intersects its own member orthogonally.
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The orthogonal trejections of the family of parabola’s y2 = 4ax + 4a2 is the familya)of circlesb)of ellipsec)itselfd)of hyper-bola’sCorrect answer is option 'C'. Can you explain this answer?
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