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

oldnick | Student

these are two hyperboles



you can write as:


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:


dividing all by 288:

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

That means:


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 | Student

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)`





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

Thus  foci is `(+-14.3178,0).`

We have another curve


we can write this as



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

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


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