Question

Find the equation for the hyperbola whose graph is shown (graph attached).

Please help and thank you in advance :)

Answer 1

i've been getting \frac{x^2}{12}-\frac{y^2}{4}=1 as the answer but I'm probably incredibly wrong...I'm doing my homework in advance so I'm not really sure how to do this one. Any help is appreciated!

Answer 2

The equation came out weird...it's supposed to be: \[\frac{ x^2 }{ 12 } - \frac{ y^2 }{ 16 } = 1\]

Answer 3

it's generally a bit trickier if you do not have the focii or directrices.

However you can do the following. You do have the major axis vertices of 2sqt(3) and the centre of (0,0)

Thus your hyperbola must fit
\[\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 } = 1 \]
When y = 0, x = 2 sqrt(3) so a = 2 sqrt(3) and a^2 = 12
\[\frac{ x^2 }{ (2 \sqrt{3})^2 } - \frac{ y^2 }{ b^2 } = 1\]
\[\frac{ x^2 }{ 12 } - \frac{ y^2 }{ b^2 } = 1\]

Then you sub in (4,4) into the equation to find the value of b (and b^2)
\[\frac{ 16 }{ 12 } - \frac{ 16 }{ b^2 } = 1\]

Answer 4

You find
\[
b = 4 \sqrt 3\\
b^2=48\\
\frac{x^2}{12}-\frac{y^2}{48}=1
\]

Answer 5

Thank you to both!!