The polar of the focus -ae-0 with respect to the hyperbola x2-a2-y2-b2-1 is

The correct option is A x= a/e Required polar is given by, T=0 ⇒ aexa^2 0b^2=1 ⇒ x= ae Hence, option 'A' is correct. ...

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Standard XII

Chemistry

Quantum Numbers

The polar of ...

Question

The polar of the focus $(- a e, 0)$ with respect to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is

x = - \frac{a}{e}

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x = \frac{a}{e}

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y = - \frac{b}{e}

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y = \frac{b}{e}

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Solution

The correct option is A $x = - \frac{a}{e}$
Required polar is given by, $T = 0$
$\Rightarrow \frac{- a e x}{a^{2}} - \frac{0}{b^{2}} = 1$
$\Rightarrow x = - \frac{a}{e}$
Hence, option 'A' is correct.

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Any chord passing through the focus (ae, 0) of the hyperbola $x^{2} - y^{2} = a^{2}$ is conjugate to the line

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