Tangents are drawn from a point $(-2\sqrt 3 ,2)$ to the hyperbola $y^2-x^2=4$ and the chord of contact subtends an angle $\theta$ at center of hyperbola. Find the value of $12 \tan^2 \theta$.
My attempt:
The equation of chord of contact is $\sqrt 3 x+y=2$. Solving it with hyperbola we get the intersection points as $(0,2)$ and $(2\sqrt 3,-4)$. So calculating the angle gives me as $\frac{\pi}{2} + \tan^{-1}{(\frac{2}{\sqrt 3})}$. Which is wrong according to answer key. Where am I wrong?
Without seeing your work, it’s impossible to tell you exactly where you’re going wrong, but let’s see:
$$\cos\theta = {(0,2)\cdot(2\sqrt3,-4) \over \lVert(0,2)\rVert \lVert(2\sqrt3,-4)\rVert} = {-8 \over 2 \cdot 2\sqrt7} = -\frac2{\sqrt7},$$ then use $\tan^2\theta+1=\sec^2\theta$ to obtain $\tan^2\theta = \frac34$. Since one of the points is on the $y$-axis, we can also compute $\tan\theta$ directly from the other point: $\tan\theta = {2\sqrt3\over-4} = -\frac{\sqrt3}2$. So, it appears that you’ve gotten a numerator and denominator swapped somewhere along the way.
As Blue noted in a comment, you don’t need to compute $\theta$ explicitly since you already have $\tan^2\theta$. Now, just multiply that by $12$.