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# Equation Of Chords Joining Two Points On Ellipse

EQUATION OF A CHORD JOINING $${\bf{\theta }}$$ and $$\phi$$:Consider two points $$\theta \;\text{and}\; \phi$$ lying on the ellipse \begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.\end{align}

We wish to determine the equation of the chord joining these two points.

Using the two point form, we have

\begin{align}&\frac{{y - b\sin \theta }}{{x - a\cos \theta }} = \frac{{b\sin \theta - b\sin \phi }}{{a\cos \theta - a\cos \phi }} \hfill \\&\Rightarrow \qquad \frac{{y - b\sin \theta }}{{x - a\cos \theta }} = \frac{{ - b}}{a}\frac{{\cos \left( {\frac{{\theta + \phi }}{2}} \right)}}{{\sin \left( {\frac{{\theta + \phi }}{2}} \right)}} \hfill \\ & \Rightarrow\qquad \frac{x}{a}\cos \left( {\frac{{\theta + \phi }}{2}} \right) + \frac{y}{b}\sin \left( {\frac{{\theta + \phi }}{2}} \right) = \cos \theta \cos \left( {\frac{{\theta + \phi }}{2}} \right) + \sin \theta \sin \left( {\frac{{\theta + \phi }}{2}} \right) \hfill \\ &\Rightarrow\quad \boxed{\frac{x}{a}\cos \left( {\frac{{\theta + \phi }}{2}} \right) + \frac{y}{b}\sin \left( {\frac{{\theta + \phi }}{2}} \right) = \cos \left( {\frac{{\theta - \phi }}{2}} \right)} \hfill \\\end{align}

This is the most general equation of a chord joining any two arbitrary points $$\theta \;and\; \phi$$ on the ellipse. As an exercise using this form try writing

(a) the equation of any chord passing through the origin

and  (b) the equation of the latus-recta by using the eccentric angles of its extremities which we derived earlier.

Example – 9

Suppose that the chord joining the points $${\theta _1}\;\text{and}\;{\theta _2}$$ on the ellipse \begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\end{align} intersects the major-axis in (h, 0). Show that

$\tan \left( {\frac{{{\theta _1}}}{2}} \right)\tan \left( {\frac{{{\theta _2}}}{2}} \right) = \frac{{h - a}}{{h + a}}.$

Solution: By the result we just derived, the equation of the chord joining $${\theta _1}\;\text{and}\;{\theta _2}$$ is

$\frac{x}{a}\cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right) + \frac{y}{b}\sin \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right) = \cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right)$

If this passes through (h, 0), we have

$\frac{{ - h}}{a}\cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right) = \cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right)$

\begin{align}&\Rightarrow\qquad\quad \frac{h}{a} = \frac{{\cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right)}}{{\cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right)}}\\&\Rightarrow \quad\; \frac{{h - a}}{{h + a}} = \frac{{\cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right) - \cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right)}}{{\cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right) + \cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2}} \right)}}\\&{\rm{ }}\qquad\qquad\qquad = \tan \left( {\frac{{{\theta _1}}}{2}} \right)\tan \left( {\frac{{{\theta _2}}}{2}} \right)\end{align}