JEE MainMathematicsModerateEllipse

The eccentricity of ellipse 14x^2- 4xy + 11y^2= 60 will be
equal to
nh?kZo`Ùk 14x^2- 4xy + 11y^2= 60

Show answer and solution

Centre of ellipse will be (0,0)

\left( {using}\frac{df}{dx} = 0\&\frac{df}{dy} = 0 \right)

Now using parametric equation from centre (0,0) any point having a
distance r from origin can be written as x = rcosθ and y= rsinθ

r^{2}\left( 1 + \cos^{2}\theta - 4sin\theta cos\theta + 11\sin^{2}\theta \right) = 60

r^{2} = \frac{60}{7(1 + cos2\theta) - 2sin2\theta + \frac{11}{2}(1 - cos2\theta)}

r^{2} = \frac{120}{3cos2\theta - 4sin2\theta + 25}

maximum value r^2 will be 6 and minimum value will be 4

a = \sqrt{6}\ and\ b = 2\ hence\ e = \sqrt{1 - \frac{4}{6}} = \frac{1}{\sqrt{3}}

Hindi. nh?kZo`Ùk Unz (0,0)

\left( {}\frac{df}{dx} = 0\&\frac{df}{dy} = 0dh\ lgk;rk\ ls \right)

pwfd izkpfyd lehdj.k Unz (0, 0) vkSj ewy fcUnq ls r nwjh
ij fuEu tk ldrk gS
x = rcosθ vkSj y = rsinθ

r^{2}\left( 1 + \cos^{2}\theta - 4sin\theta cos\theta + 11\sin^{2}\theta \right) = 60

r^{2} = \frac{60}{7(1 + cos2\theta) - 2sin2\theta + \frac{11}{2}(1 - cos2\theta)}

r^{2} = \frac{120}{3cos2\theta - 4sin2\theta + 25}

r^2 eku 6 gksxk

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx

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