Question

Prove that tangent at any point P on a parabola, bisects angle SPM, where S is focus, and M is foot of perpendicular from P on directrix.

Answer 1

The tangent at any point P on a parabola bisects the angle SPM, where S is the focus and M is the foot of the perpendicular from P to the directrix. This can be proven using coordinate geometry by setting up the parabola equation, identifying the coordinates of P, S, and M, calculating the slopes of the tangent line, SP, and PM, and then showing that the angles between the tangent and SP, and the tangent and PM are equal.

Answer 2

Let the equation of the parabola be $y^2 = 4ax$. The focus S is at $(a, 0)$ and the directrix is $x = -a$. Let P be a point $(x_0, y_0)$ on the parabola, so $y_0^2 = 4ax_0$. The foot of the perpendicular from P to the directrix is M$(-a, y_0)$.

The slope of the tangent at P is $m_t = \frac{2a}{y_0}$. The slope of SP is $m_{SP} = \frac{y_0 - 0}{x_0 - a} = \frac{y_0}{x_0 - a}$. The line segment PM is horizontal, so its slope $m_{PM} = 0$.

The angle $\alpha$ between the tangent and SP is given by: $\tan \alpha = \left| \frac{m_t - m_{SP}}{1 + m_t m_{SP}} \right| = \left| \frac{\frac{2a}{y_0} - \frac{y_0}{x_0 - a}}{1 + \frac{2a}{y_0} \cdot \frac{y_0}{x_0 - a}} \right| = \left| \frac{2a(x_0 - a) - y_0^2}{y_0(x_0 - a) + 2ay_0} \right|$ Using $y_0^2 = 4ax_0$: $\tan \alpha = \left| \frac{2ax_0 - 2a^2 - 4ax_0}{y_0x_0 - ay_0 + 2ay_0} \right| = \left| \frac{-2ax_0 - 2a^2}{y_0(x_0 + a)} \right| = \left| \frac{-2a(x_0 + a)}{y_0(x_0 + a)} \right| = \left| \frac{-2a}{y_0} \right| = \left| \frac{2a}{y_0} \right|$.

The angle $\beta$ between the tangent and PM is given by: $\tan \beta = \left| \frac{m_t - m_{PM}}{1 + m_t m_{PM}} \right| = \left| \frac{\frac{2a}{y_0} - 0}{1 + \frac{2a}{y_0} \cdot 0} \right| = \left| \frac{2a}{y_0} \right|$.

Since $\tan \alpha = \tan \beta$, the angles $\alpha$ and $\beta$ are equal, which means the tangent at P bisects $\angle SPM$.