Chapter 10.3, Problem 27E

To determine

The standard form of the equation of the ellipse whose centre $\left(1,4\right)and\text{verteces}\left(1,0\right),\left(1,8\right)anda=2c$

Answer to Problem 27E

The standard form of the equation of the ellipse is $\frac{{\left(x-1\right)}^2}{12}+\frac{{\left(y-4\right)}^2}{16}=1$

Explanation of Solution

Given information :

Given centre $\left(1,4\right)and\text{verteces}\left(1,0\right),\left(1,8\right)anda=2c$

Since value of $y$ changes for both centre $\left(h,k\right)=\left(1,4\right)\text{verteces}\left(1,0\right),\left(1,8\right)$ , so the ellipse has vertical major axis.

Vertices $\left(h,k\pm a\right)=\left(1,8\right),\left(1,0\right)$

  $\begin{array}{l}\Rightarrow 4+a=8\\ \Rightarrow a=8-4\\ \therefore a=4\end{array}$

Thus, standard equation of ellipse is $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1\mathrm{.........}(i)$

Given $a=2c$

  $\begin{array}{l}\Rightarrow 4=2c\\ \therefore c=2\end{array}$

We have $c^2=a^2+b^2$

  $\begin{array}{l}\Rightarrow 2^2=4^2-b^2\\ \Rightarrow b^2=16-4\\ \therefore b^2=12\end{array}$

Substitute $\left(h,k\right)=\left(1,4\right)a=4\text{and}{b}^2=12$ in equation $(i)$ , we get

  $\begin{array}{l}\Rightarrow \frac{{\left(x-1\right)}^2}{12}+\frac{{\left(y-4\right)}^2}{4}=1\\ \therefore \frac{{\left(x-1\right)}^2}{12}+\frac{{\left(y-4\right)}^2}{16}=1\end{array}$

Hence, the standard form of the equation of the ellipse is $\frac{{\left(x-1\right)}^2}{12}+\frac{{\left(y-4\right)}^2}{16}=1$