Answer
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Hint: The eccentricity of the ellipse is given as \[e=\dfrac{c}{a},\] so we will have to find the focal length (c). Firstly we have that latus rectum is the same as half the minor axis, i.e. \[\dfrac{2{{b}^{2}}}{a}=b\] and we will solve this.
Complete step-by-step answer:
We get the relation between b and a as a = 2b which we will use it to find c given as \[c=\sqrt{{{a}^{2}}-{{b}^{2}}}.\] Once, we have c, we will find the value of eccentricity (e).
We will consider that the equation of the ellipse is given as
\[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]
So, we have the length of the major axis as \[2\times a=2a.\]
And the length of the minor axis as \[2\times b=2b.\]
Now, we are given that our ellipse has its latus rectum equal to the half of the minor axis.
We have the latus rectum of the ellipse given as
\[\text{Latus Rectum}=\dfrac{2{{b}^{2}}}{a}\]
So, as latus rectum is equal to the half of the minor axis,
\[\Rightarrow \dfrac{2{{b}^{2}}}{a}=\dfrac{2b}{a}\]
Simplifying further, we get,
\[\Rightarrow 2{{b}^{2}}=ba\]
Cancelling the like terms, we get,
\[\Rightarrow 2b=a\]
Now, the eccentricity of the latus rectum is given as \[e=\dfrac{c}{a}\] where c is the focal length, a is the length of the semi-major axis.
So, to find e, we have to find the value of c first. We know that the focal length c is given as,
\[c=\sqrt{{{a}^{2}}-{{b}^{2}}}\]
As a = 2b. So,
\[c=\sqrt{{{\left( 2b \right)}^{2}}-{{b}^{2}}}\]
\[\Rightarrow c=\sqrt{4{{b}^{2}}-{{b}^{2}}}\]
Simplifying further, we get,
\[c=\sqrt{3{{b}^{2}}}\]
\[\Rightarrow c=b\sqrt{3}\]
As we get c as \[b\sqrt{3},\] now we will find the value of eccentricity.
\[e=\dfrac{c}{a}\]
Here, \[c=b\sqrt{3}\] and a = 2b. So, we get,
\[\Rightarrow e=\dfrac{b\sqrt{3}}{2b}\]
Cancelling the like term, we get,
\[\Rightarrow e=\dfrac{\sqrt{3}}{2}\]
So, the correct answer is “Option (c)”.
Note: Students make a simple mistake like \[\sqrt{4{{b}^{2}}-{{b}^{2}}}=\sqrt{4}\] this happens. So be careful while subtracting as the variables are not eliminated when their coefficients are not the same. Also, remember that \[2{{b}^{2}}=ba\] gives us 2b = a as b gets cancelled from both the sides. This happened because b is never zero. So, we can divide both sides by b and get 2b = a.
Complete step-by-step answer:
We get the relation between b and a as a = 2b which we will use it to find c given as \[c=\sqrt{{{a}^{2}}-{{b}^{2}}}.\] Once, we have c, we will find the value of eccentricity (e).
We will consider that the equation of the ellipse is given as
\[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]
So, we have the length of the major axis as \[2\times a=2a.\]
And the length of the minor axis as \[2\times b=2b.\]
Now, we are given that our ellipse has its latus rectum equal to the half of the minor axis.
We have the latus rectum of the ellipse given as
\[\text{Latus Rectum}=\dfrac{2{{b}^{2}}}{a}\]
So, as latus rectum is equal to the half of the minor axis,
\[\Rightarrow \dfrac{2{{b}^{2}}}{a}=\dfrac{2b}{a}\]
Simplifying further, we get,
\[\Rightarrow 2{{b}^{2}}=ba\]
Cancelling the like terms, we get,
\[\Rightarrow 2b=a\]
Now, the eccentricity of the latus rectum is given as \[e=\dfrac{c}{a}\] where c is the focal length, a is the length of the semi-major axis.
So, to find e, we have to find the value of c first. We know that the focal length c is given as,
\[c=\sqrt{{{a}^{2}}-{{b}^{2}}}\]
As a = 2b. So,
\[c=\sqrt{{{\left( 2b \right)}^{2}}-{{b}^{2}}}\]
\[\Rightarrow c=\sqrt{4{{b}^{2}}-{{b}^{2}}}\]
Simplifying further, we get,
\[c=\sqrt{3{{b}^{2}}}\]
\[\Rightarrow c=b\sqrt{3}\]
As we get c as \[b\sqrt{3},\] now we will find the value of eccentricity.
\[e=\dfrac{c}{a}\]
Here, \[c=b\sqrt{3}\] and a = 2b. So, we get,
\[\Rightarrow e=\dfrac{b\sqrt{3}}{2b}\]
Cancelling the like term, we get,
\[\Rightarrow e=\dfrac{\sqrt{3}}{2}\]
So, the correct answer is “Option (c)”.
Note: Students make a simple mistake like \[\sqrt{4{{b}^{2}}-{{b}^{2}}}=\sqrt{4}\] this happens. So be careful while subtracting as the variables are not eliminated when their coefficients are not the same. Also, remember that \[2{{b}^{2}}=ba\] gives us 2b = a as b gets cancelled from both the sides. This happened because b is never zero. So, we can divide both sides by b and get 2b = a.
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