Chapter 9.3, Problem 15E

To determine

To find the equation of the hyperbola whose vertices $\left(4,1\right),\left(4,9\right)$ and foci $\left(4,0\right),\left(4,10\right)$ .

Answer to Problem 15E

The required hyperbola equation is $\frac{{\left(y-5\right)}^2}{16}-\frac{{\left(x-4\right)}^2}{9}=1$

Explanation of Solution

Given information :

Given vertices $\left(4,1\right),\left(4,9\right)$ and foci $\left(4,0\right),\left(4,10\right)$ .

Since the $y$ co-ordinates are the ones that are changing for the vertices and foci,

So the hyperbola has a vertical transverse axis.

Thus, the required hyperbola equation is $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1\mathrm{.........}(i)$

We know that the centre is the midpoint of foci.

  $\begin{array}{l}\Rightarrow centre\left(h,k\right)=\left(\frac{4+4}{2},\frac{0+10}{2}\right)\\ =\left(\frac{8}{2},\frac{10}{2}\right)\\ \therefore \left(h,k\right)=\left(4,5\right)\\ \Rightarrow c=10-5=5anda=9-5=4\\ \Rightarrow b=\sqrt{c^2-a^2}=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}\end{array}$

Now, substitute $\left(h,k\right)=\left(4,5\right),a^2=16,b^2=9$ in equation $(i)$ , we get

  $\frac{{\left(y-5\right)}^2}{16}-\frac{{\left(x-4\right)}^2}{9}=1$

Which is the required hyperbola equation.