Chapter 8.2, Problem 66E

Solution

If the given statement is true or false.

It has been determined that the given statement is true.

Given:

The distance from a focus of an ellipse to either end point of the minor axis is half the length of the major axis.

Concept used:

The equation of an ellipse in standard form, having center $\left(0,0\right)$ , focal axis $y=0$ , foci $\left(\pm c,0\right)$ , vertices $\left(\pm a,0\right)$ , semi-major axis $a$ and semi-minor axis $b$ ; is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , where $c^2=a^2-b^2$ .

Calculation:

Consider the ellipse having the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

This ellipse has center at $\left(0,0\right)$ , focal axis $y=0$ , foci $\left(\pm c,0\right)$ , vertices $\left(\pm a,0\right)$ , semi-major axis $a$ and semi-minor axis $b$ ; where $c^2=a^2-b^2$ .

Then, foci of the ellipse are $\left(\pm c,0\right)$ and the end points of the minor axis is $\left(0,\pm b\right)$ .

The distance between the above referred focus and the end point of the minor axis is:

  $d=\sqrt{{\left(\pm c-0\right)}^2+{\left(0-\left(\pm b\right)\right)}^2}$

Simplifying,

  $d=\sqrt{{\left(\pm c\right)}^2+{\left(\mp b\right)}^2}$

On further simplification,

  $d=\sqrt{c^2+b^2}$

Put $c^2=a^2-b^2$ in the above expression to get,

  $d=\sqrt{a^2}$

Simplifying,

  $d=a$

Now, $a$ is nothing but the semi-major axis, that is, half the length of the major axis.

So, the distance from any focus of an ellipse to either end point of the minor axis is half the length of the major axis.

Hence, the given statement is true.

Conclusion:

It has been determined that the given statement is true.