What are the possible values of a for the least possible length of latus rectum?

[utkarshakash]

Homework Statement

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

Homework Equations

The Attempt at a Solution

The give equation can be rewritten as

$\left( y+b \right) ^2 = -2a\left\{ x-\dfrac{1+b^2}{2a} \right\}$

As given in question

$-b=\dfrac{1+b^2}{2a} \\<br /> 1+b^2+2ab=0$

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

 

[Saitama]

Homework Statement

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

Homework Equations

The Attempt at a Solution

The give equation can be rewritten as

$\left( y+b \right) ^2 = -2a\left\{ x-\dfrac{1+b^2}{2a} \right\}$

As given in question

$-b=\dfrac{1+b^2}{2a} \\<br /> 1+b^2+2ab=0$

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?

 

[tiny-tim]

hi utkarshakash!

$1+b^2+2ab=0$

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

have you tried completing the square?​

 

[utkarshakash]

hi utkarshakash!


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

have you tried completing the square?​

(1+b)^2+2b(a-1)=0

But how does this help?

 

[tiny-tim]

a can obviously be 1

is it possible eg for a to be 1/2 ?