Answer
Verified
394.5k+ views
Hint: 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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE