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Least possible length

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


If abscissa and ordinate of vertex of parabola [itex]y^2+2ax+2by-1=0[/itex] are equal then least possible length of latus rectum is

Homework Equations



The Attempt at a Solution


The give equation can be rewritten as

[itex]\left( y+b \right) ^2 = -2a\left\{ x-\dfrac{1+b^2}{2a} \right\}[/itex]

As given in question

[itex]-b=\dfrac{1+b^2}{2a} \\
1+b^2+2ab=0[/itex]

Length of latus rectum = a/2
But what can be the range of values of a?
 

Answers and Replies

  • #2
3,812
92

Homework Statement


If abscissa and ordinate of vertex of parabola [itex]y^2+2ax+2by-1=0[/itex] are equal then least possible length of latus rectum is

Homework Equations



The Attempt at a Solution


The give equation can be rewritten as

[itex]\left( y+b \right) ^2 = -2a\left\{ x-\dfrac{1+b^2}{2a} \right\}[/itex]

As given in question

[itex]-b=\dfrac{1+b^2}{2a} \\
1+b^2+2ab=0[/itex]

Length of latus rectum = a/2
But what can be the range of values of a?
You have "a" as a function of b, how about using that?
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,832
250
hi utkarshakash! :smile:
[itex]1+b^2+2ab=0[/itex]
well, a can't be 0, can it? :wink:

have you tried completing the square?​
 
  • #4
utkarshakash
Gold Member
855
13
hi utkarshakash! :smile:


well, a can't be 0, can it? :wink:

have you tried completing the square?​
(1+b)^2+2b(a-1)=0

But how does this help?
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
250
a can obviously be 1

is it possible eg for a to be 1/2 ?
 

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