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$3y = - 2x$ is

A.$\frac{2}{3}$

B.$\frac{1}{3}$

C.$\frac{1}{{\sqrt 3 }}$

D.None of these

Answer
Verified

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 â€˜Câ€™.

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 \[{m_1}{m_2} = \frac{{ - {b^2}}}{{{a^2}}}\].

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 â€˜Câ€™.

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 \[{m_1}{m_2} = \frac{{ - {b^2}}}{{{a^2}}}\].

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