If $\alpha, \beta$ are the two real roots of $4^{\mathrm{tn}}$ roots of unity and $\gamma, \delta$ are the…
If $\alpha, \beta$ are the two real roots of $4^{\mathrm{tn}}$ roots of unity and $\gamma, \delta$ are the other two roots of it, then the sum of the eccentricities of the conics $|z-\alpha|+|z-\beta|=4$ and $|z-\gamma|+|z-\delta|=6$ is
$\frac{5}{6}$
$\frac{5}{12}$
$\frac{3}{7}$
$\frac{4}{5}$
Solution
The real solutions of $4^{\text {th }}$ root of unity is \pm 1
$\therefore \alpha=1, \beta=-1$
and other roots are $\pm i$
$\begin{aligned}
& \gamma=i \& \delta=-i \\
& \therefore|z-1|+|z+1|=4 \\
& \Rightarrow|x+i y-1|+|x-i y+1|=4 \\
& \Rightarrow|(x-1)+i y|+|(x+1)-i y|=4 \\
& \Rightarrow \quad \sqrt{(x-1)^2+y^2}=4-\sqrt{(x+1)^2+y^2} \\
& \Rightarrow(x-1)^2+y^2=16+(x+1)^2+y^2-8 \sqrt{(x+1)^2+y^2} \\
& \Rightarrow-4 x=16-8 \sqrt{(x+1)^2+y^2} \\
& \Rightarrow 2 \sqrt{(x+1)^2+y^2}=4+x
\end{aligned}$
$\begin{aligned}
& \Rightarrow 4(x+1)^2+4 y^2=(4+x)^2 \\
& \Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1
\end{aligned}$
which is an ellipse whose eccentricity is
$\Rightarrow e_1=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
Now, $|z-\gamma|+|z+\delta|=6$
$\Rightarrow|z-i|+|z+i|=6$
$\begin{aligned}
& \Rightarrow|x+i(y-1)|+|x+i(y+1)|=6 \\
& \Rightarrow \sqrt{x^2+(y-1)^2}+\sqrt{x^2(y+1)^2}=6
\end{aligned}$
$\Rightarrow x^2+(y-1)^2=36+x^2+(y+1)^2-12 \sqrt{x^2+(y+1)^2}$
$\Rightarrow 12 \sqrt{x^2+(y+1)^2}=36+4 y$
$\Rightarrow 9\left(x^2+(y+1)^2\right)=(9+y)^2$
$\Rightarrow \frac{x^2}{8}+\frac{y^2}{9}=1$
which is an ellipse whose essentricity is:
$e_2=\sqrt{1-\frac{8}{9}}=\frac{1}{3}$
Now, $e_1+e_2=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$