Chapter 11, Problem 49RE

To determine

To find: The equation of the conic section.

Answer to Problem 49RE

The equation of the conic section is

  $\frac{4{\left(x-7\right)}^2}{225}+\frac{{\left(y-2\right)}^2}{100}=1$

Explanation of Solution

Given information:

Vertices of ellipse $v_1(7,12)\&v_2(7,-8)$ and passes through $p(1,8)$ clearly, this is an ellipse with major axis parallel to y − axis.

Calculation:

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

Here $a>b$

Length of major axis, $2a=$ Distance between vertices

  $\begin{array}{l}2a=\sqrt{{\left(7-7\right)}^2+\left(12-{\left(-8\right)}^2\right)2}\\ =\sqrt{0+{\left(20\right)}^2}\\ =2a=20\\ a=10\end{array}$

The center of ellipse, $(h,k)=\left(\frac{7+7}{2},\frac{12+\left(-8\right)}{2}\right)$

  $(h,k)=\left(7,2\right)$

Hence, Equation of ellipse is $\frac{{\left(x-7\right)}^2}{b^2}+\frac{{\left(y-2\right)}^2}{{\left(10\right)}^2}=1$

  $\frac{{\left(x-7\right)}^2}{b^2}+\frac{{\left(y-2\right)}^2}{100}=1$

Given that the ellipse passes through $p(1,8)$

  $\begin{array}{l}\frac{{\left(1-7\right)}^2}{b^2}+\frac{{\left(8-2\right)}^2}{100}=1\\ \frac{36}{b^2}+\frac{36}{100}=1\\ \frac{36}{b^2}=1-\frac{36}{100}\\ \frac{36}{b^2}=\frac{64}{100}\\ b^2=\frac{100\times 36}{64}=\frac{225}{4}\end{array}$

Hence, The equation of ellipse

  $\frac{{\left(x-7\right)}^2}{\frac{225}{4}}+\frac{{\left(y-2\right)}^2}{100}=1$

  $\frac{4{\left(x-7\right)}^2}{225}+\frac{{\left(y-2\right)}^2}{100}=1$