# Tangents from $(-2\sqrt3,2)$ to hyperbola $y^2-x^2=4$ determine a chord of contact subtending angle $\theta$ at the center. Find $12\tan^2\theta$.

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?

• What is the answer according to the key?
– Blue
Dec 15 '19 at 12:42
• Answer is 9 according to answer key. Dec 15 '19 at 12:45
• Given your value of $\theta$, what is the corresponding value of $12\tan^2\theta$?
– Blue
Dec 15 '19 at 12:46
• Moreover we want to this without a calculator. Dec 15 '19 at 12:50
• 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.)
– Blue
Dec 15 '19 at 12:52

$$\cos\theta = {(0,2)\cdot(2\sqrt3,-4) \over \lVert(0,2)\rVert \lVert(2\sqrt3,-4)\rVert} = {-8 \over 2 \cdot 2\sqrt7} = -\frac2{\sqrt7},$$ 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 $$12$$.
• 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 $(2\sqrt{3},-4)$.)