Given the ellipse with equation
Question:

Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci.

Solution:

We know that equation of an ellipse is

$=\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$

Also we have,

Length of latus rectum $=\frac{2 b^{2}}{a}$

Length of minor axis $=2 b$

$9 x^{2}+25 y^{2}=225$

Dividing the above equation by 225 ,

$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$

The above equation can be written as

$\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$

From the above equation we have,

$a=5, b=3$

$b^{2}=a^{2}\left(1-e^{2}\right)$

$3^{2}=5^{2}\left(1-e^{2}\right)$

On rearranging we get

$\frac{9}{25}=1-\mathrm{e}^{2}$

$e^{2}=1-\frac{9}{25}$

Taking LCM and simplifying we get

$e^{2}=\frac{16}{25}$

$e=\frac{4}{5}$

$3^{2}=5^{2}\left(1-e^{2}\right)$

On rearranging we get

$\frac{9}{25}=1-\mathrm{e}^{2}$

$\mathrm{e}^{2}=1-\frac{9}{25}$

Taking LCM and simplifving we get

$e^{2}=\frac{16}{25}$

$e=\frac{4}{5}$

We know foci $=(\pm a e, 0)$

$=\left(\pm 5 \times \frac{4}{5}, 0\right)=(\pm 4,0)$

Hence, the eccentricity is $4 / 5$ and foci is $(\pm 4,0)$.

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