JEE Math Practice Question
Let $$H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$$ be the hyperbola, whose eccentricity is $$\sqrt{3}$$ and the length of the latus rectum is $$4 \sqrt{3}$$. Suppose the point $$(\alpha, 6), \alpha>0$$ lies on $$H$$. If $$\beta$$ is the product of the focal distances of the point $$(\alpha, 6)$$, then $$\alpha^2+\beta$$ is equal to
- A.170
- B.171
- C.169
- D.172
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