Chapter 11, Problem 6T

To determine

To Find: The equation of a conic from the given graph.

Answer to Problem 6T

The equation of a shifted conic is ${\left(x-2\right)}^2-\frac{y^2}{3}=1$.

Explanation of Solution

Definition used: 1

“The equation of the shifted hyperbola with center as $\left(h,k\right)$ and horizontal axis (x-axis) as axis is  $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , where $a>0$ , $b>0$, $c^2=a^2+b^2$”.

Definition used: 2

“The equation of asymptotes of the shifted hyperbola $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ are

$\frac{\left(x-h\right)}{a}+\frac{\left(y-k\right)}{b}=0$ and $\frac{\left(x-h\right)}{a}-\frac{\left(y-k\right)}{b}=0$”.

Calculation:

From the given graph, it is observed that the graph is a hyperbola with horizontal axix as transverse axis and the vertices are $\left(0,1\right)$ and $\left(0,3\right)$.

Let $\left(x_1,y_1\right)$ is $\left(1,0\right)$ and $\left(x_2,y_2\right)$ is $\left(3,0\right)$.

Since the center of the hyperbola is the midpoint of the vertices, compute the midpoint.

$\begin{array}{c}M=\left(\frac{x_1+y_1}{2},\frac{x_2+y_2}{2}\right)\\ =C\left(\frac{1+3}{2},\frac{0+0}{2}\right)\\ =\left(\frac{4}{2},\frac{0}{2}\right)\\ =\left(2,0\right)\end{array}$

Therefore, the center of the ellipse is $\left(2,0\right)$.

Compute the distance between the vertices.

$\begin{array}{c}2a=2\\ a=1\end{array}$

By the above definition (1), compute the equation of the shifted ellipse.

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

Therefore, by definition (1), the equation of conic is $\frac{{\left(x-2\right)}^2}{1}-\frac{y^2}{b^2}=1$.

Since the equation $\frac{{\left(x-2\right)}^2}{1}-\frac{y^2}{b^2}=1$ passes through $\left(0,3\right)$, substitute $\left(x,y\right)=\left(0,3\right)$.

$\begin{array}{c}\frac{{\left(x-2\right)}^2}{1}-\frac{y^2}{b^2}=1\\ \frac{{\left(0-2\right)}^2}{1}-\frac{{\left(3\right)}^2}{b^2}=1\\ 4-\frac{9}{b^2}=1\\ \frac{-9}{b^2}=-3\end{array}$

Simplify the terms further,

$\begin{array}{c}{b}^2=\frac{9}{3}\\ b^2=3\\ b=\sqrt{3}\end{array}$

Substitute $b=\sqrt{3}$ in equation $\frac{{\left(x-2\right)}^2}{1}-\frac{y^2}{b^2}=1$,

$\begin{array}{c}\frac{{\left(x-2\right)}^2}{1}-\frac{y^2}{{\left(\sqrt{3}\right)}^2}=1\\ \frac{{\left(x-2\right)}^2}{1}-\frac{y^2}{3}=1\end{array}$

Therefore, the equation of conic is $\frac{{\left(x-2\right)}^2}{1}-\frac{y^2}{3}=1$.