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Let $\mathrm{X}$ - axis be the transverse axis and $\mathrm{Y}$-axis be the conjugate axis of a hyperbola $\mathrm{H}$. Let $\mathrm{x}^2+\mathrm{y}^2=16$ be the director circle of $\mathrm{H}$. If the perpendicular distance from the centre of $\mathrm{H}$ to its latus rectum is $\sqrt{34}$ then $\mathrm{a}+\mathrm{b}=$
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The correct answer is:
8
We have equation of the director circle of hyperbola
$$
\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is } x^2+y^2=a^2-b^2 \\
& \therefore a^2-b^2=16
\end{aligned}
$$
We have the perpendicular distance from centre $(0,0)$ of hyperbola to its latus rectum is ae.
$$
\begin{aligned}
& \therefore \mathrm{ae}=\mathrm{c}=\sqrt{34} \Rightarrow \mathrm{c}^2=34 \\
& \therefore \mathrm{b}^2+\mathrm{a}^2=\mathrm{c}^2=34
\end{aligned}
$$
solving equations (i) and (ii) we get $a=5 \& b=3$
$$
\therefore a+b=5+3=8 \text {. }
$$
$$
\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is } x^2+y^2=a^2-b^2 \\
& \therefore a^2-b^2=16
\end{aligned}
$$
We have the perpendicular distance from centre $(0,0)$ of hyperbola to its latus rectum is ae.
$$
\begin{aligned}
& \therefore \mathrm{ae}=\mathrm{c}=\sqrt{34} \Rightarrow \mathrm{c}^2=34 \\
& \therefore \mathrm{b}^2+\mathrm{a}^2=\mathrm{c}^2=34
\end{aligned}
$$
solving equations (i) and (ii) we get $a=5 \& b=3$
$$
\therefore a+b=5+3=8 \text {. }
$$
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