Question

Find the length of the Latus Rectum with the following equation: y= x^2 +6

Answer 1

We have the next equation

`y=x^2+6`

First, we need to find the focus of this parabola the vertice is in (0,6)

`4p\mleft(y-k\mright)=\mleft(x-h\mright)^2`

where in our case h =0, k=6

`4\cdot(1)/(4)(y-6)=x^2`

Therefore the focus will be

`(0,6+(1)/(4))=(0,(25)/(4))`

Then for Latus Rectum is located between the next points

`(-0.5,(25)/(4))\text{ and (}0.5,(24)/(5)\text{)}`

the latus Rectum

`4p=4((1)/(4))=1`

the length of the latus rectum is 1