Chapter 8, Problem 47RE

Solution

To find: the equation for the conic in the standard form.

The general equation of the of the hyperbola with center $\left(2,1\right)$ and vertices $\left(2\pm 3,1\right)$ is $\frac{{\left(x-2\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{16}=1$ .

Given data:

A hyperbola is given with center $\left(2,1\right)$ ,vertices $\left(2\pm 3,1\right)$ and asymptotes as $y=\left(\frac{4}{3}\right)\left(x-2\right)+1$ .

Formula used:

The standard equation of the hyperbola with center $\left(h,k\right)$ and vertices $\left(h\pm a,k\right)$ is given as:

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

Asymptotes : $y=\pm \frac{b}{a}\left(x-h\right)+k$

Calculation:

It is given that the vertices and center of the hyperbola is $\left(2\pm 3,1\right)$ and $\left(2,1\right)$ respectively.

  $a=3,h=2,k=1$ .

The asymptote of the ellipse is $y=\pm \frac{b}{a}\left(x-h\right)+k$ .

  $b=4$

The standard equation of the hyperbola with center $\left(2,1\right)$ and vertices $\left(2\pm 3,1\right)$ is given as:

  $\begin{array}{l}\frac{{\left(x-2\right)}^2}{3^2}-\frac{{\left(y-1\right)}^2}{4^2}=1\\ \frac{{\left(x-2\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{16}=1\end{array}$

Therefore, the general equation of the of the hyperbola with center $\left(2,1\right)$ and vertices $\left(2\pm 3,1\right)$ is $\frac{{\left(x-2\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{16}=1$ .