How do I find the foci of these? x^2/36-y^2/169=1   36x^2-8y^2=288

oldnick | (Level 1) Valedictorian

Posted on

these are two hyperboles

first:

`x^2/36-y^2/169=1`

you can write as:

`(x/6)^2-(y/13)^2=1`

so the axis are    `a=6` and `b=13`

so foci of hyperbole are   `F(-6;0);`  and `F'(6;0)`

Asimptotes (in blue) :      `y=+-13/6x`

Second hyperbole:

`36x^2-8y^2=288`

dividing all by 288:

`x^2/8 -y^2/36=1`

That means:

`(x/(2sqrt(2)))^2-(y/6)^2=1`

So `a=2sqrt(2)`   and `b=6`

so the foci are:     `F(-2sqrt(2);0)`  and   `F'(2sqrt(2);0)`

asimptotes    `y=+-3sqrt(2)/2 x`

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

Standard form of hyperbola is `x^2/a^2-y^2/b^2=1` ,

then foci are `(+-ae,0)` and `e=sqrt(1+b^2/a^2)`

`x^2/36-y^2/169=1`

`a^2=36,a=6`

`b^2=169,b=13`

`e=sqrt(1+169/36)=2.3863`

`therefore`  foci `(+-ae,0)=(+-14.3178,0)`

Thus  foci is `(+-14.3178,0).`

We have another curve

`36x^2-8y^2=288`

we can write this as

`(36x^2)/288-(8y^2)/288=1`

`x^2/8-y^2/36=1`

`therefore e=sqrt(1+36/8)=2.3452`

Thus foci are `(+-ae,0)=(+-6.6332,0)`