Find the points on the hyperbola `x ^ 2-y ^ 2 = 8`  whose distance from the center of the hyperbola equally linear eccentricity.

1 Answer | Add Yours

tiburtius's profile pic

tiburtius | High School Teacher | (Level 2) Educator

Posted on

General equation of hyperbola is

`(x-p)^2/a^2-(y-q)^2/b^2=1`

where point `C(p,q)` is center of hyperbola.

If we divide equation of our hyperbola we get

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

From this we see that center is `C(0,0)` and `a^2=b^2=8.`

Now we need to calculate linear eccentricity for which we can use the following formula

`e^2=a^2+b^2`

Hence we have

`e^2=8+8=16`

Now all points whose distance from center `C` is `e` lie on a circle with center in `C(0,0)` and with radius `e.` Since general equation of circle is

`(x-p)^2+(y-q)^2=r^2`

 equation of our circle is

`x^2+y^2=16`

Now we only need to find points that lie on both circle and hyperbola and for that we need to solve the following system of equations

`x^2-y^2=8`

`x^2+y^2=16`

If we add those two equations we get

`2x^2=24`

`x^2=12`

`x_(1,2)=pm2sqrt3`

Now we put that into second equation

`12+y^2=16`

`y_(1,2)=pm2`

So your solutions are `(-2sqrt3,-2),` `(-2sqrt3,2),` `(2sqrt3,-2)` and   `(2sqrt3,2).`

We’ve answered 318,957 questions. We can answer yours, too.

Ask a question