The locus of the point (asecθ+btanθ,atanθ+bsecθ)where 0≤θ<2π is

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The locus of the point $(a \sec θ + b \tan θ, a \tan θ + b \sec θ)$ where $0 \leq θ < 2 π$ is

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${(x^{2} - y^{2})}^{2} = 16 x y$

$x^{2} - y^{2} = a^{2} - b^{2}$

$x^{2} + y^{2} = a^{2} + b^{2}$

$x^{2} - y^{2} = a^{2} + b^{2}$

answer is D.

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Detailed Solution

Let the point be (x, y)
$(x, y) = (a \sec θ + b Tan θ, aTan θ + b \sec θ)$
$\Rightarrow x = a \sec θ + b Tan θ, y = aTan θ + b \sec θ$
$\begin{array}{l} = x^{2} - y^{2} = a^{2} \sec^{2} θ + b^{2} {Tan}^{2} θ + 2 a b \sec θ \tan θ - a^{2} {Tan}^{2} θ - \\ b^{2} \sec^{2} θ - 2 a b Tan θ \sec θ \end{array} x^{2} - y^{2} = a^{2} (1) - b^{2} (1) = a^{2} - b^{2}$