Question

Can you please help what I wrote out is what the teacher gave us(Find the standard equation of a hyperbola given: vertices:

Answer 1

As given by the question

There are given that the vertices and focii point is:

`\begin{gathered} \text{vertices: (}\pm6,\text{ 0)} \\ Foci\colon\text{ (}\pm\text{8, 0)} \end{gathered}`

Now,

From the general form of the equation of hyperbola:

`((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1`

Then,

Firs find the center (h, k) of the hyperbola:

So

We know that the center of hyperbola must be at the midpoint of the given foci:

Then,

The point of foci is:

`(8,\text{ 0) and (-8, 0)}`

So,

The center of the hyperbola is:

`\begin{gathered} (h,\text{ k)=(}(8+(-8))/(2),\text{ }(0+0)/(2)) \\ (h,\text{ k)=(}(8-8)/(2),\text{ }(0+0)/(2)) \\ (h,\text{ k)=(0, 0)} \end{gathered}`

Now,

Find a which equals half the length of the vertical major axis.

The distance of any focus of the hyperbola from the center is denoted c

So,

The distance of focus (8, 0) and the center (0,0) is:

`\begin{gathered} c=\sqrt[]{(0-8)^2+(0-0)^2} \\ c=\sqrt[]{(8)^2+(0)^2} \\ c=\sqrt[]{64+0} \\ c=\sqrt[]{64} \\ c=8 \end{gathered}`

Now,

The distance of any vertex of the hyperbola from the center is denoted a

So,

The vertex point is:

`(-6,\text{ 0) and (6, 0)}`

Then,

`\begin{gathered} b=\sqrt[]{(0-6)^2+(0-0)^2} \\ b=\sqrt[]{36} \\ b=6 \end{gathered}`

And,

Find a which equals half the length of the horizontal minor axis and we know that the relationship between c, a and b.

So,

From the formula of retaionship between c, a and b.

`\begin{gathered} c^2=a^2+b^2 \\ 8^2=a^2+6^2 \\ 64=a^2+36 \\ a^2=64-36 \\ a^2=28 \end{gathered}`

Then,

From the standard form of the hyperbola:

`\begin{gathered} ((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1 \\ \frac{(x-0)^2}{28^{}}+\frac{(y-0)^2}{36^{}}=1 \end{gathered}`

Hence, the standard form of the hyperbola is shown below:

`\frac{(x-0)^2}{28^{}}+\frac{(y-0)^2}{36^{}}=1`