Reflective property of the ellipse | Analytic Geometry | Mathematics

Theorem 5.2 (Reflective property of the ellipse). A ray leaving one focus of an ellipse is reflected by the ellipse through the other focus.

Proof. Take the ellipse of Theorem 4.1 with \(c > 0\) (if \(c = 0\) it is a circle, and every ray from the centre returns along itself). Let \(P(x_1, y_1)\) be on the ellipse.

If \(y_1 = 0\), then \(P = (\pm a, 0)\), the tangent there is vertical, and the ray runs along the \(x\)-axis, so it returns along the axis through both foci. Suppose now \(y_1 \neq 0\) and, for the moment, \(x_1 \neq \pm c\), so that neither focal line is vertical.

Slopes. Differentiating \(x^2/a^2 + y^2/b^2 = 1\) implicitly gives \(2x/a^2 + (2y/b^2)\,y' = 0\). So the tangent at \(P\) has slope

\[ m = -\frac {b^2 x_1}{a^2 y_1}, \]

and the lines \(F_1P\) and \(F_2P\) have slopes

\[ m_1 = \frac {y_1}{x_1 + c}, \qquad m_2 = \frac {y_1}{x_1 - c}. \]

Below we use \(b^2 x_1^2 + a^2 y_1^2 = a^2 b^2\) (because \(P\) is on the ellipse) and \(a^2 - b^2 = c^2\).

Incoming angle. Let \(\alpha \) be the angle from \(F_1P\) to the tangent. By Theorem 5.1, \(\tan \alpha = (m - m_1)/(1 + m m_1)\). Multiply the numerator and the denominator by \(a^2 y_1 (x_1 + c)\):

\[ \begin {aligned} \text {numerator} &= -b^2 x_1 (x_1 + c) - a^2 y_1^2 = -(b^2 x_1^2 + a^2 y_1^2) - b^2 c x_1 = -b^2 (a^2 + c x_1), \\ \text {denominator} &= a^2 y_1 (x_1 + c) - b^2 x_1 y_1 = y_1 \bigl ( (a^2 - b^2) x_1 + a^2 c \bigr ) = c y_1 (a^2 + c x_1). \end {aligned} \]

Since \(\abs {x_1} \leq a\) and \(c < a\), we have \(a^2 + c x_1 > 0\). So the denominator is not zero (the lines are not perpendicular), and

\[ \tan \alpha = -\frac {b^2}{c y_1}. \]

Outgoing angle. Let \(\beta \) be the angle from the tangent to \(F_2P\), so \(\tan \beta = (m_2 - m)/(1 + m m_2)\). Multiply the numerator and the denominator by \(a^2 y_1 (x_1 - c)\):

\[ \begin {aligned} \text {numerator} &= a^2 y_1^2 + b^2 x_1 (x_1 - c) = b^2 (a^2 - c x_1), \\ \text {denominator} &= a^2 y_1 (x_1 - c) - b^2 x_1 y_1 = -c y_1 (a^2 - c x_1), \end {aligned} \]

with \(a^2 - c x_1 > 0\). Hence

\[ \tan \beta = -\frac {b^2}{c y_1} = \tan \alpha . \]

Both angles lie in \([0, \pi )\), where \(\tan \) is one-to-one, so \(\alpha = \beta \). So \(F_1P\) and \(F_2P\) make equal angles with the tangent, one on each side: reflecting the line \(F_1P\) in the tangent gives the line \(F_2P\). This is the law of reflection (the angle of incidence equals the angle of reflection), so the reflected ray passes through \(F_2\).

When \(x_1 = \pm c\), one focal line is vertical and the slope formula does not apply. The result still holds there by continuity: both angles depend continuously on \(P\), and they agree at every nearby point of the ellipse.