Ellipse

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If $y = x$ and $3 y + 2 x = 0$ are the equations
of a pair of conjugate diameters of an ellipse, then the
eccentricity of the ellipse is

Easy

Solution

: Let the equation of the ellipse be
$x^{2} / a^{2} + y^{2} / b^{2} = 1$
Slope of the given diameters are $m_{1} = 1, m_{2} = - 2 / 3$
$\Rightarrow m_{1} m_{2} = - 2 / 3 = - b^{2} / a^{2}$
[using the condition of conjugacy of two diameters]
$\begin{array}{l} 3 b^{2} = 2 a^{2} \Rightarrow 3 a^{2} (1 - e^{2}) = 2 a^{2} \\ 1 - e^{2} = 2 / 3 \Rightarrow e^{2} = 1 / 3 \Rightarrow e = 1 / \sqrt{3} \end{array}$

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Ellipse

Remember concepts with our Masterclasses.

If y = x and 3y+2x=0 are the equations of a pair of conjugate diameters of an ellipse, then the eccentricity of the ellipse is

: Let the equation of the ellipse be x2/a2+y2/b2=1 Slope of the given diameters are m1=1,m2=−2/3 ⇒m1m2=−2/3=−b2/a2[using the condition of conjugacy of two diameters] 3b2=2a2⇒3a21−e2=2a21−e2=2/3⇒e2=1/3⇒e=1/3

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If $y = x$ and $3 y + 2 x = 0$ are the equations
of a pair of conjugate diameters of an ellipse, then the
eccentricity of the ellipse is