The pole of the line $2 x + 4 y + 5 = 0$ with respect to the parabola $y^{2} = 8 a x$ , is

(\frac{5}{2}, - 8 a)

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(\frac{5}{4}, - 5 a)

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(\frac{5}{4}, - 2 a)

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none of the these

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Solution

The correct option is D $(\frac{5}{2}, - 8 a)$
Let the pole be $(x_{1}, y_{1})$ . Then, the equation of the polar with respect to $y^{2} = 8 a x$ is
$y y_{1} = 4 a (x + x_{1})$
or, $4 a x - y y_{1} + 4 a x_{1} = 0$ ...(i)
But line (i) and $2 x + 4 y + 5 = 0$ represent the same line.
$∴ \frac{4 a}{2} = - \frac{y_{1}}{4} = \frac{4 a x_{1}}{5}$

$⟹ x_{1} = \frac{5}{2}$ and $y_{1} = - \frac{16 a}{2}$
Hence, the pole of the given line is $(\frac{5}{2}, - 8 a)$

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