Question

If the lines joining the focii of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)  where a>b and an extremity of its minor axis is inclined at an angle 60°, then the eccentricity of the ellipse is

• A. \(\frac{\sqrt3}{2}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{\sqrt7}{3}\)
• D. \(\frac{1}{\sqrt3}\)

Answer 1

The Correct Option is B

Given:

Ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a > b\)

Foci: \((\pm ae, 0)\) 

Extremity of minor axis: \((0, b)\)

The lines joining the foci to the extremity of the minor axis are inclined at an angle of 60°. This implies the angle between the line joining (ae,0) and (0,b) to the x-axis is 30 degrees since 2 such lines create an isosceles triangle.

We want to find the eccentricity (e) of the ellipse.

Let's consider the focus (ae, 0) and the extremity of the minor axis (0, b). The slope of the line joining these two points is:

\(m = \frac{b - 0}{0 - ae} = -\frac{b}{ae}\)

The angle \(\theta\) between this line and the x-axis satisfies:

\(\tan \theta = |m| = \frac{b}{ae}\)

We are given that the angle between the line joining the focus and the endpoint of the minor axis is 60 degrees. Since the triangle formed by the focii and endpoint of minor axis is isoceles, the angle the line joining the focus to the x-axis is (180 - 60)/2 = 60 degrees. Therefore the other two angles are 60 degrees too.

So, \(\tan(60^\circ) = \sqrt{3} = \frac{b}{ae}\), which means \(b = ae\sqrt{3}\)

We know that the relationship between a, b, and e in an ellipse is:

\(b^2 = a^2(1 - e^2)\)

Substitute \(b = ae\sqrt{3}\) into this equation:

\((ae\sqrt{3})^2 = a^2(1 - e^2)\)

\(3a^2e^2 = a^2 - a^2e^2\)

\(3e^2 = 1 - e^2\)

\(4e^2 = 1\)

\(e^2 = \frac{1}{4}\)

\(e = \frac{1}{2}\) (Since eccentricity must be positive)

Therefore, the eccentricity of the ellipse is \(\frac{1}{2}\).