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If $x = h + a \sec θ$ and $y = k + b cosec θ$ , then

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$\frac{a^{2}}{{(x + h)}^{2}} - \frac{b^{2}}{{(y + k)}^{2}} = 1$

$\frac{a^{2}}{{(x - h)}^{2}} + \frac{b^{2}}{{(y - k)}^{2}} = 1$

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$

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

answer is B.

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Detailed Solution

$\frac{x - h}{a} = \sec θ \Rightarrow \cos θ = \frac{a}{x - h} \dots \dots \dots (1)$

$\frac{y - k}{b} = cosec θ \Rightarrow \sin θ = \frac{b}{y - k} \dots \dots \dots (2)$

$\frac{a^{2}}{{(x - h)}^{2}} + \frac{b^{2}}{{(y - k)}^{2}} = \cos^{2} θ + \sin^{2} θ = 1$

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