If the chord joining the points `(a sectheta_1, b tantheta_1)` and `(a sectheta_2, b tantheta_2)` on the hyperbola `x^2/a^2-y^2/b^2=1` is a focal chord, then prove that `tan(theta_1/2)tan(theta_2/2)+(ke-1)/(ke+1)=0`, where `k=+-1`

Options

A. `(1-e)/(1+e)`
B. `(e-1)/(e+1)`
C. `(e+1)/(e-1)`
D. `(1+e)/(1-e)`

Explanation

To prove that if the chord joining the points \((a \sec \theta_1, b \tan \theta_1)\) and \((a \sec \theta_2, b \tan \theta_2)\) on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is a focal chord, then \( \tan\left(\frac{\theta_1}{2}\right) \tan\left(\frac{\theta_2}{2}\right) + \frac{ke - 1}{ke + 1} = 0 \), where \( k = \pm 1 \).

### Step-by-Step Solution

1. **Identify the Focus of the Hyperbola:**
The foci of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are located at \( (ae, 0) \) and \( (-ae, 0) \), where \( e = \sqrt{1 + \frac{b^2}{a^2}} \).

**Hint:** Remember that the eccentricity \( e \) is defined in terms of the semi-major and semi-minor axes.
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