Question

One of the focus of ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ is $\left( {0,5\sqrt 3 } \right)$ and difference in lengths of major and minor axis is $5$ units. Then length of latus rectum is
A. $3$
B. $5$
C. $10$
D. $15$

Answer 1

In this problem, we will consider $\left( {0,5\sqrt 3 } \right) = \left( {0,be} \right)$ because $x$ coordinate is zero. Also given that the difference of $b$ (length of major axis) and $a$ (length of minor axis) is $5$ units. By using the formula ${b^2}{e^2} = {b^2} - {a^2}$, we will find the sum of $b$ (length of major axis) and $a$ (length of minor axis). The length of the latus rectum is obtained by using the formula $\dfrac{{2{a^2}}}{b}$.

Complete step-by-step solution
In this problem, the equation of an ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ and one of the focus is $\left( {0,5\sqrt 3 } \right)$.
As $x$ the coordinate of the focus is zero, we can say that the focus is lying on the $Y$ axis. Also we can say that $b > a$ where $b$ is the length of major axis and $a$ is the length of minor axis.
In this case, we can say that the focus is $\left( {0,be} \right)$ where $e$ is the eccentricity of ellipse. Let us compare $\left( {0,5\sqrt 3 } \right)$ with $\left( {0,be} \right)$. Therefore, we get $be = 5\sqrt 3 \cdots \cdots \left( 1 \right)$.
Squaring on both sides of equation $\left( 1 \right)$, we get ${b^2}{e^2} = 25 \times 3 = 75 \cdots \cdots \left( 2 \right)$.
Also given that the difference of $b$ (length of major axis) and $a$ (length of minor axis) is $5$ units. Therefore, we can write $b - a = 5 \cdots \cdots \left( 3 \right)$.

Now we will use the formula ${b^2}{e^2} = {b^2} - {a^2}$ to find the sum of $b$ (length of major axis) and $a$ (length of minor axis). Therefore, we get
${b^2}{e^2} = {b^2} - {a^2} \\\ \Rightarrow {b^2}{e^2} = \left( {b - a} \right)\left( {b + a} \right)\quad \left[ {\because {a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right] \\\ \Rightarrow 75 = 5\left( {b + a} \right)\quad \left[ {\because {b^2}{e^2} = 75,\;b - a = 5} \right] \\\ \Rightarrow b + a = \dfrac{{75}}{5} \\\ \Rightarrow b + a = 15 \cdots \cdots \left( 4 \right) \\\$
Adding equation $\left( 3 \right)$ and $\left( 4 \right)$, we get
$b - a + b + a = 5 + 15 \\\ \Rightarrow 2b = 20 \\\ \Rightarrow b = \dfrac{{20}}{2} \\\ \Rightarrow b = 10 \\\$
Let us substitute $b = 10$ in equation $\left( 4 \right)$, we get
$10 + a = 15 \\\ \Rightarrow a = 15 - 10 \\\ \Rightarrow a = 5 \\\$
Now we will find the length of latus rectum (LR) by using the formula $\dfrac{{2{a^2}}}{b}$. Therefore, we get
length of latus rectum (LR) $= \dfrac{{2{{\left( 5 \right)}^2}}}{{10}} = \dfrac{{2 \times 25}}{{10}} = 5$. Therefore, option B is true.

Note: $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ is the standard form of an ellipse. For example, let us take $\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1$. In this equation, we can see that denominator of the term $\dfrac{{{x^2}}}{9}$ is larger than the denominator of the term $\dfrac{{{y^2}}}{4}$. In this case, we will compare the equation $\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1$ with $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ to find the eccentricity and foci. The eccentricity of an ellipse $\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1$ is obtained by using the formula $e = \dfrac{{\sqrt {{a^2} - {b^2}} }}{a}$ where $a > b$ and the foci are $\left( { \pm ae,0} \right)$ where $e$ is the eccentricity.