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If $\theta$ is the acute angle between the tangents drawn from the point $(2,3)$ to the hyperbola $5 x^2-6 y^2-30=0$, then $\tan \theta=$
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Verified Answer
The correct answer is:
$\frac{4}{3}$
Given equation of ellipse $\frac{x^2}{6}-\frac{y^2}{5}=1$
Let equation of tangent $y=m x+c$
$\therefore$ It passes through $(2,3)$ i.e $c=3-2 m$
By condition of tangency.
$$
\begin{aligned}
& c^2=a^2 m^2-b^2 \Rightarrow c^2=6 m^2-5 \\
& \Rightarrow m^2+6 m-7=0 \quad \text { (from (i)) } \\
& \Rightarrow m=1,-7
\end{aligned}
$$
Now, $\tan \theta=\left|\frac{1+7}{1-7}\right|=\frac{8}{6}=\frac{4}{3}$.
Let equation of tangent $y=m x+c$
$\therefore$ It passes through $(2,3)$ i.e $c=3-2 m$
By condition of tangency.
$$
\begin{aligned}
& c^2=a^2 m^2-b^2 \Rightarrow c^2=6 m^2-5 \\
& \Rightarrow m^2+6 m-7=0 \quad \text { (from (i)) } \\
& \Rightarrow m=1,-7
\end{aligned}
$$
Now, $\tan \theta=\left|\frac{1+7}{1-7}\right|=\frac{8}{6}=\frac{4}{3}$.
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