The centre of an ellipse is at (0, 0), major axis is on the y-axis. If the ellipse passes through (3, 2) and (1, 6), then what is its eccentricity ?

Concept:

The equation of ellipse with centre (0,0) is of the form, \({x^2 \over a^2 } +{y^2 \over b^2} = 1\)

For an ellipse \({x^2 \over a^2 } +{y^2 \over b^2} = 1\) with y-axis as major axis, the eccentricity = \(e = \sqrt{1 - {a^2 \over b^2}}\)

Calculation:

Given: The centre of an ellipse is at (0, 0), major axis is on the y-axis and the ellipse passes through (3, 2) and (1, 6).

Let the equation of the ellipse be \({x^2 \over a^2 } +{y^2 \over b^2} = 1\)

(3,2) and (1,6) lies on the ellipse

⇒ \({3^2 \over a^2 } +{2^2 \over b^2} = 1\) and \({1^2 \over a^2 } +{6^2 \over b^2} = 1\)

⇒ \({9 \over a^2 } +{4 \over b^2} = 1\) __(i)

and \({1 \over a^2 } +{36 \over b^2} = 1\) __(ii)

Subtracting (i) from (ii),

⇒ \({1-9 \over a^2 } +{36 -4 \over b^2} = 0\)

 ⇒ 32a²​ = 8b²

⇒ 4a² = b²

Put in (i) 

⇒ \({9 \over a^2 } +{4 \over 4a^2} = 1\)

⇒ a² = 10

⇒ b² = 4 × 10 = 40 

So, the equation of the ellipse will be \({x^2 \over 10} +{y^2 \over 40} = 1\)

Now for the ellipse \({x^2 \over 10} +{y^2 \over 40} = 1\), the eccentricity of the ellipse is given by,

 \(e = \sqrt{1 - {10 \over 40}}\)

⇒ \(e = \sqrt{1 - {1 \over 4}}= \sqrt{ {3 \over 4}}\)

⇒ Eccentricity = \(\frac{\sqrt{3}}{2}\)

∴ The correct option is (1).