The equation of a line having slope $- \tan θ$ and passing through the point $(a^{2} \sin θ, a^{2} \cos θ)$ is

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$x \cos θ + y \sin θ = a$

$x \sin θ + y \cos θ = a^{2}$

$x \cos θ + y \sin θ = a^{2}$

$x \sin θ + y \cos θ = a$

answer is C.

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Detailed Solution

Slope $(m) = - \tan θ$

Let $(x_{1}, y_{1}) = (a^{2} \sin θ, a^{2} \cos θ)$

$y - y_{1} = m (x - x_{1})$

$y - a^{2} \cos θ = - \tan θ (x - a^{2} \sin θ)$

$y - a^{2} \cos θ = \frac{- \sin θ}{\cos θ} (x - a^{2} \sin θ)$

$y \cos θ - a^{2} \cos^{2} θ = - x \sin θ + a^{2} \sin^{2} θ$

$x \sin θ + y \cos θ = a^{2} \sin^{2} θ + a^{2} \cos^{2} θ$

$x \sin θ + y \cos θ = a^{2} (1)$

$x \sin θ + y \cos θ = a^{2}$

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