The locus of the poles of tangents to the auxiliary circle with respect to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , is

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{1}{a^{2}}

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\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{b^{2}}

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\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}}

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Solution

The correct option is C $\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}}$
let $(h, k)$ be the pole
Then equation of polar is $\frac{h x}{a^{2}} + \frac{k y}{b^{2}} = 1$
Since, it is tangent to $x^{2} + y^{2} = a^{2}$
Then $a = p$
$\Rightarrow a = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \frac{1}{\sqrt{\frac{h^{2}}{a^{4}} + \frac{k^{2}}{b^{4}}}} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
Therefore, locus is $\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}}$

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