Chapter 9, Problem 36RE

To determine

Calculate the standard form of the equation of the ellipse.

Answer to Problem 36RE

The standard from of the equation is $\frac{x^2}{\frac{9}{2}}+\frac{y^2}{36}=1$ .

Explanation of Solution

Given:

It is given in the question that the $vertices:(0,\pm 6)$ ; passes through the point $(2,2)$ .

Concept Used:

In this, use the concept of the ellipse and use the equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Calculation: Firstly, calculate the length of the major axis. This equals to half the distance between the two given vertices.

So, $b=6$

Now, the vertices lie on the vertical axis of the ellipse. The center of the ellipse is the midpoint of the vertices, so the center is at $(0,0)$ . plug this information and the given point on the ellipse to find $a^2$ , which is the square of half the minor axis (horizontal in this case) length.

So,

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{{(2)}^2}{a^2}+\frac{{(2)}^2}{{(6)}^2}=1\\ \frac{4}{a^2}=1-\frac{4}{36}\\ \frac{4}{a^2}=\frac{32}{36}\\ a^2=\frac{36\times 4}{32}\\ a^2=\frac{9}{2}\end{array}$

Put the above value in the equation,

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{x^2}{\frac{9}{2}}+\frac{y^2}{{(6)}^2}=1\\ \frac{x^2}{\frac{9}{2}}+\frac{y^2}{36}=1\end{array}$

Conclusion:

The standard equation is $\frac{x^2}{\frac{9}{2}}+\frac{y^2}{36}=1$ .