Answer to Question #122649 in Analytic Geometry for Nikhil

Question

In a parabola show that the tangent at any point makes equal angles with the foci radius of the point and the line parallel to the x-axis through the point

Answer 1

We begin with a parabola and choose the coordinate system so that the equation takes the form $x^2=4cy, c>0.$ We can express $y$ in terms of $x$ by writing

$$y={x^2\over 4c}$$

Since $\dfrac{dy}{dx}=\dfrac{x}{2c},$ the tangent line at the point $P(x_0,y_0)$ has slope $m=\dfrac{x_0}{2c}$ and has equation 

$$y-y_0=\dfrac{x_0}{2c}(x-x_0)$$

The focus is $F(0, c).$

Since the tangent line at $P(x_0, y_0)$ is not vertical, it intersects the y-axis at some point $T.$ To find the coordinates of T, we set $x=0$ in the equation of the tangent line

$$y-y_0=\dfrac{x_0}{2c}(0-x_0)$$

$$y=y_0-\dfrac{x_0^2}{2c},\ y_0=\dfrac{x_0^2}{4c},$$

$$y=-y_0=-\dfrac{x_0^2}{4c}$$

We have point $T(0,-y_0)$ and $d(F, T)=c+y_0.$

Find the distance $d(F, P)$

$$d(F,P)=\sqrt{(x_0-0)^2+(y_0-c)^2}=$$

$$=\sqrt{4cy_0+y_0^2-2y_0c+c^2}=\sqrt{(y_0+c)^2=}$$

$$=|y_0+c|=y_0+c=d(F,T)\ (c>0, y_0>)$$

Since $d(F,P)=d(F,T),$ the triangle $TFP$ is isosceles and the angles marked $\alpha$ and $\beta$ are equal. Since a ray $l$ is parallel to the y-axis, $\alpha=\gamma$ and thus $\beta=\gamma).$

This means that light emitted from a source at the focus of a parabolic mirror is reflected in a beam parallel to the axis of that mirror; this is the principle of the searchlight.