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# If $\left( 2,4 \right)$ and $\left( 10,10 \right)$ are the ends of a latus-rectum of an ellipse with eccentricity $\dfrac{1}{2}$, then the length of semi-major axis is(a) $\dfrac{20}{3}$(b) $\dfrac{15}{3}$(c) $\dfrac{40}{3}$(d) None of these

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Hint: Find the length of latus rectum using distance formula between its end points and compare it with the formula of length of latus rectum. Form another equation relating the parameters $a,b$ with the eccentricity of the ellipse. Solve the two equations to find the value of the semi-major axis.

We have $\left( 2,4 \right)$ and $\left( 10,10 \right)$ as ends of latus rectum of an ellipse with eccentricity $e=\dfrac{1}{2}$. We have to find the length of the semi-major axis of this ellipse.
Let’s assume that the ellipse is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. The eccentricity of this ellipse is $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}=\dfrac{1}{2}$.
Simplifying the above equation, we have $1-\dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{1}{4}$.
\begin{align} & \Rightarrow \dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{3}{4} \\ & \Rightarrow {{b}^{2}}=\dfrac{3}{4}{{a}^{2}}.....\left( 1 \right) \\ \end{align}
We will now find the length of the latus rectum using the distance formula between ends of the latus rectum.
We know that distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$.
Substituting ${{x}_{1}}=2,{{y}_{1}}=4,{{x}_{2}}=10,{{y}_{2}}=10$ in the above equation, we have length of latus rectum as $\sqrt{{{\left( 2-10 \right)}^{2}}+{{\left( 4-10 \right)}^{2}}}=\sqrt{{{\left( -8 \right)}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{64+36}=\sqrt{100}=10$ units.
We know the length of the latus rectum has the formula $\dfrac{2{{b}^{2}}}{a}$. Thus, we have $\dfrac{2{{b}^{2}}}{a}=10$.
Simplifying the above equation, we have${{b}^{2}}=5a.....\left( 2 \right)$.
Using equations $\left( 1 \right),\left( 2 \right)$ we have $5a=\dfrac{3}{4}{{a}^{2}}$.
Simplifying the above equation by rearranging the terms, we have $a\left( \dfrac{3}{4}a-5 \right)=0$.
$\Rightarrow a=0,\dfrac{20}{3}$
As $a=0$ is not possible, we have $a=\dfrac{20}{3}$.
We know that the length of the semi-major axis is half of the length of the major axis, which is $2a$. Thus, the length of the semi-major axis is $a=\dfrac{20}{3}$ units.
Hence, the length of semi-major axis of the given ellipse is $\dfrac{20}{3}$ which is option (a).

Note: We can also solve this question by directly writing the formula for length of latus rectum in terms of eccentricity of the ellipse and then solve it to get the length of semi-major axis.
Last updated date: 17th Sep 2023
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