Question

Which is the equation of a hyperbola with directrices at x = ±2 and foci at (6, 0) and (−6, 0)?

Answer 1

`\displaystyle (x^2)/(12)-(y^2)/(24)=1`

Explanation:

Since the directrices are vertical lines:

• Use the hyperbola equation
  `\displaystyle ((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1`
• Foci are
  `(\pm c,0)`
• 
  `a^2+b^2=c^2`
• 
  `a` and
  `b` are half the lengths of the tranverse and conjugate axes respectively
• Directrices are located at the lines
  `\displaystyle x=\pm(a^2)/(c)`
• We assume the center to be
  `(h,k)\rightarrow(0,0)`

Since we already know our directrices are at the lines
`x=\pm2` and we know that
`c=\pm6` from our foci, we can solve for the value of
`a^2`:

`x=(a^2)/(c)\\\\2=(a^2)/(6)\\\\12=a^2`

This allows us to solve for the value of
`b^2`:

`a^2+b^2=c^2\\\\12+b^2=6^2\\\\12+b^2=36\\\\b^2=24`

Thus, our equation for the hyperbola is
`\displaystyle (x^2)/(12)-(y^2)/(24)=1`.