Conic Sections - Eccentricity (Ellipse)

Find the eccentricity of the ellipse given by the equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get $a^2 = 25$ and $b^2 = 9$. Thus, $a = 5$ and $b = 3$. The eccentricity is calculated as: $e = \sqrt{1 - \frac{b^2}{a^2}}$ $e = \sqrt{1 - \frac{9}{25}}$ $e = \sqrt{\frac{25 - 9}{25}}$ $e = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0.8$

We identify the semi-major and semi-minor axes from the denominator of the standard form and substitute them into the eccentricity formula.

An ellipse has its foci at $(\pm 4, 0)$ and its semi-major axis is $5$. Find its eccentricity and the equation of the ellipse.

Given foci at $(\pm ae, 0) = (\pm 4, 0)$, we have $ae = 4$. Given $a = 5$. $e = \frac{ae}{a} = \frac{4}{5} = 0.8$ To find $b^2$: $b^2 = a^2(1 - e^2)$ $b^2 = 25\left(1 - \left(\frac{4}{5}\right)^2\right)$ $b^2 = 25\left(1 - \frac{16}{25}\right)$ $b^2 = 25 \times \frac{9}{25} = 9$ The equation is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

We use the coordinates of the foci to find the value of $c$ (which is $ae$), then use $a$ to find $e$ and subsequently calculate $b^2$ to write the equation.

Calculate the distance between the foci if $a = 10$ and $e = 0.6$.

The distance between the foci is $2c$ or $2ae$. $\text{Distance} = 2 \times a \times e$ $\text{Distance} = 2 \times 10 \times 0.6$ $\text{Distance} = 12$

The foci are located at $(\pm ae, 0)$. The distance between them is the absolute difference between $ae$ and $-ae$, which is $2ae$.