If $ heta$ 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 $ an heta=$

Question

If $	heta$ 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 $	an 	heta=$

Aicharya · Accepted Answer

Given equation of ellipse $\frac{x^2}{6}-\frac{y^2}{5}=1$ Let equation of tangent $y=m x+c$ $	herefore$ 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 	ext { (from (i)) } \ & \Rightarrow m=1,-7 \end{aligned} $ Now, $	an 	heta=\left|\frac{1+7}{1-7}ight|=\frac{8}{6}=\frac{4}{3}$

If $\theta$ is the acute angle between the tangents drawn from the point $(2,3)$ to the hyperbola $5 x^2-6…

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