. . Reflective property of parabola The light or sound or radio waves originating at a parabola’s focus are reflected parallel to the parabola’s axis (Fig. .
) and conversely the rays arriving parallel to the axis are directed towards the focus (Fig. . ). Example .
The equation y = models cross sections of parabolic mirrors that are used for solar energy. There is a heating tube located at the focus of each parabola; how high is this tube located above the vertex of the parabola? Equation of the parabola is y = x That is x = y ; the vertex is ( , ) = ( ) y ⇒ a = So the heating tube needs to be placed at focus ( , ) a . Hence the heating tube needs to be placed units above the vertex of the parabola.
Example . A search light has a parabolic reflector (has a cross section that forms a ‘bowl’). The parabolic bowl is cm wide from rim to rim and cm deep. The bulb is located at the focus .
( ) What is the equation of the parabola used for reflector? ( ) How far from the vertex is the bulb to be placed so that the maximum distance covered? Fig. .
Fig. . P ( , y ) S ( , ) P' X Y S O ◄ ◄ ◄ ◄ ◄ ◄ ◄ ◄ - - Two Dimensional Analytical Geometry - II Let the vertex be ( , ) . The equation of the parabola is y = ax ( ) Since the diameter is cm and the depth is cm , the point lies on the parabola.
= a × a = Equation is y = x . ( ) The bulb is at focus ( , ) a . Hence the bulb is at a distance of cm from the vertex. Example .
An equation of the elliptical part of an optical lens system is x = . The parabolic part of the system has a focus in common with the right focus of the ellipse .The vertex of the parabola is at the origin and the parabola opens to the right. Determine the equation of the parabola. In the given ellipse a = , b = then c = a c = = c = ± Therefore the foci are F F