Question

The equation of ellipse for which, distance between directrices is $\dfrac{{25}}{2}$ and the minor axis is 6, is
A. $\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1$
B. $\dfrac{{{x^2}}}{{225}} + \dfrac{{{y^2}}}{9} = 16$
C. $\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{{25}} = 1$
D. $\dfrac{{{x^2}}}{{625}} + \dfrac{{{y^2}}}{{81}} = 1$

Answer 1

First let the equations of ellipse be of the form $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ or $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$. Next find the value of $a$ in terms of $e$ using the formula of length of directrix, which is $\dfrac{{2a}}{e}$, where $a$ is half the major the axis and $e$ is the eccentricity of the ellipse. Next, use the given length of minor axis $b$ and the condition ${b^2} = {a^2}\left( {1 - {e^2}} \right)$ to find the value of $e$. Substitute the value of $e$ to find the value of $a$ and hence the equation of ellipse.

Complete step by step solution:
Let the equation of the ellipse are $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ or $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$
We are given that the distance between two directrices is $\dfrac{{25}}{2}$.
Then, the length of the directrix is $\dfrac{{2a}}{e}$, where $a$ is half the major the axis and $e$ is the eccentricity of the ellipse.
Therefore,
$\dfrac{{25}}{2} = \dfrac{{2a}}{e} \\\ \Rightarrow \dfrac{a}{e} = \dfrac{{25}}{4} \\\$
$a = \dfrac{{25}}{4}e$
Also, we know that ${b^2} = {a^2}\left( {1 - {e^2}} \right)$, where $b$ is half of the minor axis, $a$ is half the major the axis and $e$ is the eccentricity of the ellipse.
We are given that 6 is the length of the major axis, then $b = 3$
This implies,
${3^2} = {a^2} - {a^2}{e^2} \\\ \Rightarrow 9 = {a^2} - {a^2}{e^2} \\\$
On substituting the value of $a$ we will get,
$9 = {\left( {\dfrac{{25}}{4}e} \right)^2} - {\left( {\dfrac{{25}}{4}e} \right)^2}{e^2} \\\ \Rightarrow 9 = \dfrac{{625{e^2} - 625{e^4}}}{{16}} \\\ \Rightarrow 625{e^4} - 625{e^2} + 144 = 0 \\\$
Factorise the above equation.
$625{e^4} - 400{e^2} - 225{e^2} + 144 = 0 \\\ \Rightarrow 25{e^2}\left( {25{e^2} - 16} \right) - 9\left( {25{e^2} - 16} \right) = 0 \\\ \Rightarrow \left( {25{e^2} - 9} \right)\left( {25{e^2} - 16} \right) = 0 \\\$
Equate each factor to 0 to find the value of $e$
$\left( {25{e^2} - 9} \right) = 0 \\\ \Rightarrow {e^2} = \dfrac{9}{{25}} \\\ \Rightarrow e = \dfrac{3}{5} \\\$
$\left( {25{e^2} - 16} \right) = 0 \\\ \Rightarrow {e^2} = \dfrac{{16}}{{25}} \\\ \Rightarrow e = \dfrac{4}{5} \\\$
When $e = \dfrac{3}{5}$, then
$a = \dfrac{{25}}{4}\left( {\dfrac{3}{5}} \right) \\\ a = \dfrac{{15}}{4} \\\$
Then, the equation of ellipse will be
$\dfrac{{{x^2}}}{{{{\left( {\dfrac{{15}}{4}} \right)}^2}}} + \dfrac{{{y^2}}}{{{{\left( 3 \right)}^2}}} = 1 \\\ \Rightarrow \dfrac{{16{x^2}}}{{225}} + \dfrac{{{y^2}}}{9} = 1 \\\$
Or
$\dfrac{{{x^2}}}{{{{\left( 3 \right)}^2}}} + \dfrac{{{y^2}}}{{{{\left( {\dfrac{{15}}{4}} \right)}^2}}} = 1 \\\ \Rightarrow \dfrac{{{x^2}}}{9} + \dfrac{{16{y^2}}}{{225}} = 1 \\\$
If $e = \dfrac{4}{5}$, then
$a = \dfrac{{25}}{4}\left( {\dfrac{4}{5}} \right) \\\ a = 5 \\\$
Then, the equation of the ellipse is
$\dfrac{{{x^2}}}{{{{\left( 5 \right)}^2}}} + \dfrac{{{y^2}}}{{{{\left( 3 \right)}^2}}} = 1 \\\ \Rightarrow \dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1 \\\$
But, major axis could be $y$ axis and minor could be $y$ axis.
Then the equation of ellipse will also be
$\dfrac{{{x^2}}}{{{{\left( 3 \right)}^2}}} + \dfrac{{{y^2}}}{{{5^2}}} = 1 \\\ \Rightarrow \dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{{25}} = 1 \\\$

Hence, option A and C are correct.

Note:
The standard equation of the ellipses are $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ or $\dfrac{{{x^2}}}{{{b^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$, where $a > b$.
When the denominator corresponding to $x$ is greater, then the major axis is along the $x$ axis and if denominator corresponding to the $y$ axis is the major axis.