Question

If the distance between the foci of an ellipse is 6 and the length of the minor axis is 8, then the eccentricity is :

Answer 1

Answer:

Given an ellipse with eccentricity 0.8 and a line parallel to the major axis bisecting the semi-minor axis, we want to determine the angle between the major axis and the line joining the point of intersection of the parallel line and the ellipse and the centre of the ellipse.

$Let \: the \: ellipse \: be \: \frac{x {}^{2} }{a {}^{2} } + \frac{y {}^{2} }{b {}^{2} } = 1.$

Let us assume with out loss of generality that the centre of the ellipse is at the origin and that the major axis is along the X axis.

Eccentricity

$e= \sqrt{1 - \frac{b {}^{2} }{a {}^{2} } } \: \: ⇒ \sqrt{1 - \frac{b {}^{2} }{a {}^{2} } } = 0.8.$

$⇒1− \frac{b {}^{2} }{a {}^{2} } = 0.64 \: ⇒ \: \frac{b {}^{2} }{a {}^{2} } \: = 0.36$

$⇒b=0.6a.$

$Let a=5ka=5k units ⇒b=3k⇒b=3k units.$

⇒ The distance between the line bisecting the semi-minor axis and the X axis is 1.5k units.

⇒ The Y coordinate of the point of intersection of the line bisecting the semi-minor axis and the ellipse is 1.5k units.

Since this point lies on the ellipse it must satisfy the equation of the ellipse.

$⇒ \frac{x {}^{2} }{(5k) {}^{2} } + \frac{(1.5k) {}^{2} }{(3k) {}^{2} } = 1⇒$

$\frac{x {}^{2} }{25k {}^{2} } = 1 - \frac{2.25k {}^{2} }{9k {}^{2} } = 1 - \frac{1}{4} = \frac{3}{4}$

$⇒x {}^{2} = \frac{75k {}^{2} }{4}⇒x± (\frac{5 \sqrt{3} }{2} )k.$

⇒ The coordinates of the point of intersection of the line bisecting the semi-minor axis and the ellipse are

$( \frac{5 \sqrt{3k} }{2} \: \frac{3k}{ {}^{} 2} ) \: and \: ( - \frac{5 \sqrt{3k} }{2} \: \frac{3k}{2} ).$

⇒ The required angles are arctan

${}^{ \:arctan } \: {}^{} (\frac{3}{5 \sqrt{3} }) and  {}^{arctan} ( - \frac{3}{5 \sqrt{3} } )$

⇒ The required angles are approximately 19.11o and 160.89o.

It may be noted that we have considered the angles with respect to the positive X axis.