Chapter 9.3, Problem 18E

To determine

To find the equation of the hyperbola whose vertices $\left(-2,1\right),\left(2,1\right)$ and passes through the point $\left(5,4\right)$ .

Answer to Problem 18E

The required hyperbola equation is $\frac{{\left(x\right)}^2}{4}-\frac{7{\left(y-1\right)}^2}{12}=1$

Explanation of Solution

Given information :

Given vertices $\left(-2,1\right),\left(2,1\right)$ and passes through the point $\left(5,4\right)$ .

Since the $x$ co-ordinates are the ones that are changing for the vertices,

So the hyperbola has a horizontal transverse axis.

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

Since vertices are $\left(-2,1\right),\left(2,1\right)\Rightarrow centre\left(h,k\right)=\left(0,1\right)$ .

  $\Rightarrow a=2$

Now, substitute $\left(h,k\right)=\left(0,1\right),a^2=4$ in equation $(i)$ , we get

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

The above equation passes through the point $\left(5,4\right)$ .

  $\Rightarrow \frac{{\left(5\right)}^2}{4}-\frac{{\left(4-1\right)}^2}{b^2}=1$

  $\begin{array}{l}\Rightarrow \frac{25}{4}-\frac{9}{b^2}=1\\ \Rightarrow \frac{9}{b^2}=\frac{25}{4}-1\\ \Rightarrow \frac{9}{b^2}=\frac{21}{4}\\ \Rightarrow b^2=\frac{4}{21}\times 9\\ \therefore b^2=\frac{12}{7}\end{array}$

Hence, the required hyperbola equation is $\frac{{\left(x\right)}^2}{4}-\frac{{\left(y-1\right)}^2}{\raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=1or\frac{{\left(x\right)}^2}{4}-\frac{7{\left(y-1\right)}^2}{12}=1$