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\[{F_2}B\] is a right angle, then the square of the eccentricity of the ellipse is

A) \[\dfrac{1}{2}\]

B) \[\dfrac{1}{{\sqrt 2 }}\]

C) \[\dfrac{1}{{2\sqrt 2 }}\]

D) \[\dfrac{1}{4}\]

Answer
Verified

If two lines are perpendicular then the product of their slopes is equal to -1.

The eccentricity for an ellipse with equation \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] is given by:-

\[{e^2} = 1 - \dfrac{{{b^2}}}{{{a^2}}}\]……………………………..(1)

Since it is given that \[{F_1}\] and \[{F_2}\] are foci of ellipse

Hence the coordinates of \[{F_1}\] and \[{F_2}\] are \[{F_1} \equiv \left( {ae,0} \right)\] and \[{F_2} \equiv \left( { - ae,0} \right)\]

Also OB is the minor axis therefore the coordinate of B is \[B \equiv \left( {0,b} \right)\]

Now we will find the slope of line \[{F_1}B\].

The slope m of a line passing through two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by:-

\[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]

Hence the slope \[{m_1}\] of \[{F_1}B\] having endpoints \[{F_1} \equiv \left( {ae,0} \right)\]and \[B \equiv \left( {0,b} \right)\] is given by:-

\[{m_1} = \dfrac{{b - 0}}{{0 - ae}}\]

Solving it further we get:-

\[{m_1} = \dfrac{{ - b}}{{ae}}\]…………………………(2)

Now we will find the slope of line \[{F_2}B\].

The slope m of a line passing through two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by:-

\[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]

Hence the slope \[{m_2}\] of \[{F_2}B\] having endpoints \[{F_2} \equiv \left( { - ae,0} \right)\]and \[B \equiv \left( {0,b} \right)\] is given by:-

\[{m_2} = \dfrac{{b - 0}}{{0 - \left( { - ae} \right)}}\]

Solving it further we get:-

\[{m_2} = \dfrac{b}{{ae}}\]……………………………………..(3)

Now we know that if two lines are perpendicular then the product of their slopes is equal to -1.

Now we are given that the angle between \[{F_1}B\] and \[{F_2}B\] is a right angle

Therefore \[{F_1}B\] and \[{F_2}B\] are perpendicular to each other therefore the product of their slopes will be -1.

Hence multiplying equation 2 and equation 3 we get:-

\[{m_1} \times {m_2} = - 1\]

Putting in the values from equation 2 and 3 we get:-

\[\dfrac{{ - b}}{{ae}} \times \dfrac{b}{{ae}} = - 1\]

Solving it further we get:-

\[

\dfrac{{ - {b^2}}}{{{a^2}{e^2}}} = - 1 \\

\dfrac{{{b^2}}}{{{a^2}}} = {e^2}...........................\left( 4 \right) \\

\]

Now from equation 1 we get:-

\[

{e^2} = 1 - \dfrac{{{b^2}}}{{{a^2}}} \\

\Rightarrow \dfrac{{{b^2}}}{{{a^2}}} = 1 - {e^2} \\

\]

Putting this value in equation 4 we get:-

\[1 - {e^2} = {e^2}\]

Solving it further we get:-

\[

2{e^2} = 1 \\

\Rightarrow {e^2} = \dfrac{1}{2} \\

\]

The eccentricity for an ellipse with equation \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] is given by:-

\[{e^2} = 1 - \dfrac{{{b^2}}}{{{a^2}}}\] for \[a > b\]

If \[a < b\] then eccentricity is given by:-

\[{e^2} = 1 - \dfrac{{{a^2}}}{{{b^2}}}\]