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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?

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  • $\begingroup$ What is the answer according to the key? $\endgroup$
    – Blue
    Dec 15 '19 at 12:42
  • $\begingroup$ Answer is 9 according to answer key. $\endgroup$ Dec 15 '19 at 12:45
  • $\begingroup$ Given your value of $\theta$, what is the corresponding value of $12\tan^2\theta$? $\endgroup$
    – Blue
    Dec 15 '19 at 12:46
  • $\begingroup$ Moreover we want to this without a calculator. $\endgroup$ Dec 15 '19 at 12:50
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    $\begingroup$ Your calculation involves an inverse tangent, so taking the tangent isn't really a calculator exercise. (Indeed, you didn't really even have to find $\theta$ itself. You just need the value of $\tan\theta$ to complete the problem.) $\endgroup$
    – Blue
    Dec 15 '19 at 12:52
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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 $12$.

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    $\begingroup$ OP's numerator and denominator are swapped because the $\pi/2$ term effectively turns the inverse tangent expression into an inverse cotangent (and changes a sign). (BTW: since one ray of the angle coincides with the $y$-axis, the angle's trig values are easily calculated from the coordinates of the point $(2\sqrt{3},-4)$.) $\endgroup$
    – Blue
    Dec 15 '19 at 21:01

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