Question

If the eclipse with the equation $9{x^2} + 25{y^2} = 225$ , then find the eccentricity and foci of the eclipse.

Answer 1

Firstly, simplify the equation $9{x^2} + 25{y^2} = 225$ by writing it in the form of $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ and thus find the values of a and b.
Then, to find the eccentricity of the eclipse use the formula ${b^2} = {a^2}\left( {1 - {e^2}} \right)$ and find e.
Finally, foci of any eclipse are given by foci $\equiv \left( { \pm ae,0} \right)$ . Thus, find the focus of the eclipse.

Complete step by step solution:
The given equation of the eclipse is $9{x^2} + 25{y^2} = 225$ .
Now, we will write the given equation in the form of $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ as follows
$9{x^2} + 25{y^2} = 225$
$\Rightarrow \dfrac{{9{x^2}}}{{225}} + \dfrac{{25{y^2}}}{{225}} = 1 \\\ \Rightarrow \dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1 \\\ \Rightarrow \dfrac{{{x^2}}}{{{5^2}}} + \dfrac{{{y^2}}}{{{3^2}}} = 1 \\\$
On comparing $\dfrac{{{x^2}}}{{{5^2}}} + \dfrac{{{y^2}}}{{{3^2}}} = 1$ by $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ , we get $a = 5$ and $b{\text{ = }}3$ .

Now, eccentricity of the eclipse is given by the formula ${b^2} = {a^2}\left( {1 - {e^2}} \right)$ .
$\Rightarrow {3^2} = {5^2}\left( {1 - {e^2}} \right) \\\ \Rightarrow 9 = 25\left( {1 - {e^2}} \right) \\\ \Rightarrow \dfrac{9}{{25}} = 1 - {e^2} \\\ \Rightarrow \dfrac{9}{{25}} - 1 = - {e^2} \\\ \Rightarrow \dfrac{{9 - 25}}{{25}} = - {e^2} \\\ \Rightarrow - \dfrac{{16}}{{25}} = - {e^2} \\\ \Rightarrow {e^2} = \dfrac{{16}}{{25}} \\\ \Rightarrow e = \sqrt {\dfrac{{16}}{{25}}} \\\ \Rightarrow e = \dfrac{4}{5} \\\$
Thus, the eccentricity of the eclipse with the equation $9{x^2} + 25{y^2} = 225$ is $e = \dfrac{4}{5}$ .
Now, foci of an ellipse with eccentricity e is given by $\left( { \pm ae,0} \right)$ .
Thus, foci of ellipse with $e = \dfrac{4}{5}$ and $a = 5$ is
Foci $\equiv \left( { \pm 5 \times \dfrac{4}{5},0} \right)$
$\equiv \left( { \pm 4,0} \right)$

Thus, we get the eccentricity and foci of ellipse $9{x^2} + 25{y^2} = 225$ are $e = \dfrac{4}{5}$ and $\left( { \pm 4,0} \right)$ respectively.

Note:
Eclipse:
An eclipse is a plane curve surrounding two focal points, such that for every point on the curve, the sum of the two distances of the points and the focal points is always constant.
Circle is a special case of eclipse where both the foci points lie on each other.
The general equation of eclipse is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ , where $a \geqslant b$ and the foci are $\left( { \pm c,0} \right)$ for $c = \sqrt {{a^2} - {b^2}}$ .