Question

What is the equation for the hyperbola shown? PLEASE HELP

Answer 1

ANSWER

$\frac{ {y}^{2} }{ 25} - \frac{ {x}^{2} }{ 64} = 1$

EXPLANATION

The given hyperbola has a vertical transverse axis and its center is at the origin.

The standard equation of such a parabola is:

$\frac{ {y}^{2} }{ {a}^{2} } - \frac{ {x}^{2} }{ {b}^{2} } = 1$

Where 2a=10 is the length of the transverse axis and 2b=16 is the length of the conjugate axis.

This implies that

$a = 5 \: \: and \: \: b = 8$

Hence the required equation of the hyperbola is:

$\frac{ {y}^{2} }{ {5}^{2} } - \frac{ {x}^{2} }{ {8}^{2} } = 1$

This simplifies to,

$\frac{ {y}^{2} }{ 25} - \frac{ {x}^{2} }{ 64} = 1$

Answer 2

Answer:

$\frac{(y^2}{25}-\frac{x^2}{64}=1$

Step-by-step explanation:

We have been given an image of a hyperbola. We are asked to write an equation for our given hyperbola.

We can see that our given hyperbola is a vertical hyperbola as it opens upwards and downwards.

We know that equation of a vertical hyperbola is in form $\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$, where, $(h,k)$ represents center of hyperbola.

'a' is vertex of hyperbola and 'b' is co-vertex.

We can see that center of parabola is at origin (0,0).

We can see that vertex of parabola is at point $(0,5)\text{ and }(0,-5)$, so value of a is 5.

We can see that co-vertex of parabola is at point $(8,0)\text{ and }(-8,0)$, so value of b is 8.

$\frac{(y-0)^2}{5^2}-\frac{(x-0)^2}{8^2}=1$

Therefore, our required equation would be $\frac{(y^2}{25}-\frac{x^2}{64}=1$.