Find Focii and Directrices of (x-1)(y-2) = 2

[ancient aspen]

pls help 🙂

[ancient aspen]

<@&286206848099549185>

[queen estuary]

hi do you need help

[ancient aspen]

Yes please

[upper viper]

do uk that this curve is a rectangular hyperbola

if yes then the focii are at croot2 distance from origin

standard equation is xy = c^2

and directrix is just the opp

[ancient aspen]

yes

but

This isnt a standard rectangular hyperbola

[upper viper]

so?

it is just shifted

[ancient aspen]

I see, so I just shift my origin and solve?

[upper viper]

yes

[ancient aspen]

So focii would be 3,4 and -1,0

And directrices?

[wary valve]

!nosols

[plucky saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[ancient aspen]

Sorry

[upper viper]

do uk the definition of a directrix

[ancient aspen]

I was the one who asked the question breh xD but anyways

Yes

[wary valve]

Oops you are OP. Disregard

[ancient aspen]

I was thinking directrix would have slope perpendicular to axis

and axis has slope 1

so directrix would be x+y=lambda

[upper viper]

yes

now directrix has a specific distance

a/e from origin

oh ur a jee kid

lol

but i hope u got the point

[ancient aspen]

xD what

How did you know?

[upper viper]

ur about me

[ancient aspen]

Ah I see

xD

aight anyways

[upper viper]

yeah

[ancient aspen]

So if I make distance of 3,4 from directrix equal to |ae - a/e|, then I should get lambda

[upper viper]

yes

or

u could use origin

[ancient aspen]

So x+y = 5?

[upper viper]

yeah whatever

im not doing calcs

im just giving the idea

[ancient aspen]

aight

Thank youuu!

Friends?

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