# Least possible length

1. Oct 21, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
If abscissa and ordinate of vertex of parabola $y^2+2ax+2by-1=0$ are equal then least possible length of latus rectum is

2. Relevant equations

3. 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

$-b=\dfrac{1+b^2}{2a} \\ 1+b^2+2ab=0$

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

2. Oct 21, 2013

### Saitama

You have "a" as a function of b, how about using that?

3. Oct 21, 2013

### tiny-tim

hi utkarshakash!
well, a can't be 0, can it?

have you tried completing the square?​

4. Oct 22, 2013

### utkarshakash

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

But how does this help?

5. Oct 22, 2013

### tiny-tim

a can obviously be 1

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