Chapter 9, Problem 5CT

To determine

To find: Equation of Ellipse.

Answer to Problem 5CT

  $\frac{{\left(x+6\right)}^2}{16}+\frac{{\left(y-3\right)}^2}{49}=1$

Explanation of Solution

From given graph

We can say that

Equation of major axis is $x=-6$

Equation of minor axis is $y=3$

Intersection of major axis and minor axis gives the center of Ellipse.

Therefore

Center of ellipse = $\left(h,k\right)=\left(-6,3\right)$

We know length of major axis = $2a$

End points of Major axis are $\left(-6,10\right)and\left(-6,-4\right)$

Therefore

Length of Major axis = $\sqrt{{\left(-6-\left(-6\right)\right)}^2+{\left(10-\left(-4\right)\right)}^2}$

  $\begin{array}{l}\Rightarrow 2a=\sqrt{{\left(-6+6\right)}^2+{\left(10+4\right)}^2}\\ \Rightarrow 2a=\sqrt{{\left(0\right)}^2+{\left(14\right)}^2}\\ \Rightarrow 2a=\sqrt{{\left(14\right)}^2}\\ \Rightarrow 2a=14\\ \Rightarrow a=\frac{14}{2}\\ \Rightarrow a=7\end{array}$

We know length of Minor axis is $2b$ .

End points of Minor axis are $\left(-10,3\right)and\left(-2,3\right)$

  $\begin{array}{l}\Rightarrow 2b=\sqrt{{\left(-10-\left(-2\right)\right)}^2+{\left(3-3\right)}^2}\\ \Rightarrow 2b=\sqrt{{\left(-10+2\right)}^2+{\left(0\right)}^2}\\ \Rightarrow 2b=\sqrt{{\left(-10+2\right)}^2}\\ \Rightarrow 2b=\sqrt{{\left(-8\right)}^2}\\ \Rightarrow 2b=\sqrt{64}\\ \Rightarrow 2b=8\\ \Rightarrow 2b=8\\ \Rightarrow b=\frac{8}{2}\\ \Rightarrow b=4\end{array}$

The standard form of the ellipse with center $\left(h,k\right)$ and major axis parallel to $y$ -axis and with length of semi-major axis $a$ and length of semi-minor axis $b$ is

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

Therefore, the required equation of ellipse is,

  $\frac{{\left(x-\left(-6\right)\right)}^2}{4^2}+\frac{{\left(y-3\right)}^2}{7^2}=1$

  $\Rightarrow \frac{{\left(x+6\right)}^2}{16}+\frac{{\left(y-3\right)}^2}{49}=1$