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# The eccentricity of an ellipse whose pair of a conjugate diameter are $y = x$ and $3y = - 2x$ isA.$\frac{2}{3}$B.$\frac{1}{3}$C.$\frac{1}{{\sqrt 3 }}$D.None of these

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Hint: Here, we will use the conjugate diameters concept of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$.

Given, the conjugate diameters of the ellipse are
$y = x$$\to (1) and 3y = - 2x$$ \to (2)$
Let us compare equation (1) with $y = {m_1}x$ such that we can obtain ${m_1} = 1$ and equation (2) with c$y = {m_2}x$ and we can obtain ${m_2} = \frac{{ - 2}}{3}$.
As we know, if the conjugate diameters of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ are in the form of $y = {m_1}x$ and
$y = {m_2}x$ then
${m_1}{m_2} = \frac{{ - {b^2}}}{{{a^2}}} \to (3)$i.e.., the product of their slopes will be equal
to $\frac{{ - {b^2}}}{{{a^2}}}$.
Now, let us substitute the values of ${m_1}$ and ${m_2}$ in the equation$(3)$, we get
$\Rightarrow (1)(\frac{{ - 2}}{3}) = - \frac{{{b^2}}}{{{a^2}}}$
$\Rightarrow 2{a^2} = 3{b^2} \\ \Rightarrow 2{a^2} = 3{a^2}(1 - {e^2})[\because {b^2} = {a^2}(1 - {e^2})] \\ \Rightarrow 2 = 3(1 - {e^2}) \\ \Rightarrow {e^2} = \frac{1}{3} \\ \Rightarrow e = \frac{1}{{\sqrt 3 }} \\$
So the eccentricity obtained is $\frac{1}{{\sqrt 3 }}$ .
Hence the correct option is ‘C’.

Note: In ellipse, two diameters are said to be conjugate when each bisects all chords parallel
to each other. Two diameters of form $y = {m_1}x$ and $y = {m_2}x$ are said to be
conjugate if ${m_1}{m_2} = \frac{{ - {b^2}}}{{{a^2}}}$.

Last updated date: 01st Oct 2023
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