Ellipse
10.0 Chord of contact
10.0 Chord of contact
The equation of chord of contact of tangents drawn from an external point $P({x_1},{y_1})$ to the
ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ can be find out using $T=0$ i.e., $$\frac{{x{x_1}}}{{{a^2}}} + \frac{{y{y_1}}}{{{b^2}}} = 1$$
Question 13. If the eccentric angles of the ends of a focal chord of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1{\text{ }}(a > b)$ be $\alpha $ and $\beta $, prove that $$\tan \frac{\alpha }{2}.\tan \frac{\beta }{2} = \frac{{e - 1}}{{e + 1}}{\text{ or }}\frac{{e + 1}}{{e - 1}}$$
Solution: As we know that the equation of chord drawn from two points with eccentric angles $\alpha $ and $\beta $ is $$\frac{x}{a}\cos \frac{{\alpha + \beta }}{2} + \frac{y}{b}\sin \frac{{\alpha + \beta }}{2} = \cos \frac{{\alpha - \beta }}{2}$$
Since, It passes through the focus $(ae,0)$.
Therefore,
$$\begin{equation} \begin{aligned} e\cos \frac{{\alpha + \beta }}{2} = \cos \frac{{\alpha - \beta }}{2} \\ {\text{Using Trigonometric identities, on expanding we get}} \\ e[\cos \frac{\alpha }{2}\cos \frac{\beta }{2} - \sin \frac{\alpha }{2}\sin \frac{\beta }{2}] = cos\frac{\alpha }{2}\cos \frac{\beta }{2} + \sin \frac{\alpha }{2}\sin \frac{\beta }{2} \\ {\text{Taking cos}}\frac{\alpha }{2}\cos \frac{\beta }{2}{\text{ common from both the sides, we get}} \\ e[1 - \tan \frac{\alpha }{2}\tan \frac{\beta }{2}] = 1 + \tan \frac{\alpha }{2}\tan \frac{\beta }{2} \\ e - 1 = \tan \frac{\alpha }{2}\tan \frac{\beta }{2}(e + 1) \\ \tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{{e - 1}}{{e + 1}} \\\end{aligned} \end{equation} $$
Similarly, If the chord passes through $(-ae,0)$, then $$\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{{e + 1}}{{e - 1}}$$
