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If $x \cos α + y \sin α = x \cos β + y \sin β = 2 a,$ then $\cos α \cos β$ is,

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a

\frac{4 a x}{x^{2} + y^{2}}

b

\frac{4 a^{2} - y^{2}}{x^{2} + y^{2}}

c

\frac{4 a y}{x^{2} + y^{2}}

d

\frac{4 a^{2} - x^{2}}{x^{2} + y^{2}}

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detailed solution

Correct option is B

We have,

$x \cos α + y \sin α = x \cos β + y \sin β = 2 a$

$\Rightarrow α, β$ are the roots of $x \cos θ + y \sin θ = 2 a .$

Now,

$\begin{array}{l} x \cos θ + y \sin θ = 2 a \\ \Rightarrow (x \cos θ - 2 a)^{2} = y^{2} \sin^{2} θ \\ \Rightarrow x^{2} \cos^{2} θ - 4 a x \cos θ + 4 a^{2} = y^{2} (1 - \cos^{2} θ) \\ \Rightarrow (x^{2} + y^{2}) \cos^{2} θ - 4 a x \cos θ + 4 a^{2} - y^{2} = 0 \end{array}$

Clearly, $\cos α, \cos β$ are the roots of this equation.

$∴ \cos α \cos β = \frac{4 a^{2} - y^{2}}{x^{2} + y^{2}}$ .

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