If α is a variable complex number such that |α|>1 and z=α+1α lies on a conic then

Let |α|=r>1 and α rcisθ thenz=x+iy=α+1α=rcisθ+cis(−θ)t ⇒x=(r+1r)cosθ and y=(r−1r)sinθ Eliminating θ gives x2(r+1r)2+y2(r−1r)=1 Which is an ellipse. a=r+1r,b=r−1r(r=|α|>1⇒a>b) ∴ 1−b2a2e=2r+1r, distance between foci =2ae=4distance between directrices =2ae=z1+z22

If α is a variable complex number such that |α|>1 and z=α+1α lies on a conic then

Let |α|=r>1 and α rcisθ thenz=x+iy=α+1α=rcisθ+cis(−θ)t ⇒x=(r+1r)cosθ and y=(r−1r)sinθ Eliminating θ gives x2(r+1r)2+y2(r−1r)=1 Which is an ellipse. a=r+1r,b=r−1r(r=|α|>1⇒a>b) ∴ 1−b2a2e=2r+1r, distance between foci =2ae=4distance between directrices =2ae=z1+z22

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If $α$ is a variable complex number such that $| α | > 1$ and $z = α + \frac{1}{α}$ lies on a conic then

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Length of latus rectum is $\frac{2 (| α | - 1)}{{| α |}^{2} + 1}$

Eccentricity of the conic is $\frac{2 | α |}{1 + {| α |}^{2}}$

Distance between foci is 4

Distance between directrices is ${(| α | + \frac{1}{| α |})}^{2}$

answer is A, B, D.

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Detailed Solution

Let $| α | = r > 1 and α r c i s θ t h e n z = x + i y = α + \frac{1}{α} = r c i s θ + \frac{c i s (- θ)}{t}$
$\Rightarrow x = (r + \frac{1}{r}) \cos θ and y = (r - \frac{1}{r}) \sin θ$
Eliminating $θ$ gives $\frac{x^{2}}{{(r + \frac{1}{r})}^{2}} + \frac{y^{2}}{(r - \frac{1}{r})} = 1$
Which is an ellipse. $a = r + \frac{1}{r}, b = r - \frac{1}{r} (r = | α | > 1 \Rightarrow a > b)$
$∴ \sqrt[e]{1 - \frac{b^{2}}{a^{2}}} = \frac{2}{r + \frac{1}{r}},$ distance between foci =2ae=4
distance between directrices $= \frac{2 a}{e} = \frac{z_{1} + z_{2}}{2}$