$α$ and $β$ are the eccentric angles of the extremities of a focal chord of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , then $\frac{{cos}^{2} (\frac{α - β}{2})}{{cos}^{2} (\frac{α + β}{2})} =$

\frac{a^{2} + b^{2}}{a^{2}}

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\frac{a^{2}}{a^{2} + b^{2}}

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\frac{a^{2}}{a^{2} - b^{2}}

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\frac{a^{2} - b^{2}}{a^{2}}

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Solution

The correct option is B $\frac{a^{2} - b^{2}}{a^{2}}$
$(α, β)$ are eccentric angles of extremities of focal chord.
Equation of chord with $(α, β)$ as extremities is $\frac{x}{a} cos (\frac{α + β}{2}) + \frac{y}{b} sin (\frac{α + β}{2}) = cos (\frac{α - β}{2})$
Since, it passes through $(a e, 0)$
$e cos (\frac{α + β}{2}) = cos (\frac{α - β}{2})$

$\frac{{cos}^{2} (\frac{α - β}{2})}{{cos}^{2} (\frac{α + β}{2})} = e^{2} = \frac{a^{2} - b^{2}}{a^{2}}$

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Similar questions

If A = [\begin{matrix} a & b b & a \end{matrix}] and A^{2} = [\begin{matrix} α & β β & α \end{matrix}], then

A = [\begin{matrix} a & b b & a \end{matrix}]

A^{2} = [\begin{matrix} α & β β & α \end{matrix}]

A = [\begin{matrix} a & b b & a \end{matrix}]

A^{2} = [\begin{matrix} α & β β & α \end{matrix}]

If $A = [\begin{matrix} a & b b & a \end{matrix}]$ and $A^{2} = [\begin{matrix} α & β β & α \end{matrix}]$ , then

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

x^{2} - y^{2} = 0

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