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If the reflector is 80 cm in diameter and 40 cm deep, how far from the opening is the light source?

this is the complete problem: A searchlight reflector is designed so that a cross section through it axis is a parabola with the light source at its focus. If the reflector is 80 cm in diameter and 40 cm deep, how far from the opening is the light source?: A searchlight reflector is designed so that a cross section through it axis is a parabola with the light source at its focus. If the reflector is 80 cm in diameter and 40 cm deep, how far from the opening is the light source?

~ please show your solution to provide me a better understanding about this parabola problem. Thank you very much.

Set coordinate system with origin in vertex of the parabola and y axis going through focus, that has coordinates (0, F).
Use definition of parabola that states that parabola is place of points that has equal distance to focus and directrix. So vertex is in between focus and directrix. So directrix formula is y = -F.
The edge of the light source has coordinates (40, 40) and is equal distance to directrix (F + 40) and to focus sqrt(40^2 + (40 -F)^2). So we can write (40 + F)^2 = 40^2 + (40 - F)^2, or 40^2 = (40 + F)^2 - (40 - F)^2 = 4*40*F
F = 10
Then light source is 40 - F = 30 cm from opening.

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please help?

1.what would you do to make right the lateral inversion by a plane mirror?( hint: think about multiple reflections)

2.why does a mirror invert the image side ways but not upside down?

3.can you modify a periscope to see the objects behind you? how will you do it?

4.can we make a kaleidoscope using four mirror surfaces? will it make better or worse images than the three mirror one?

5. if two mirrors are kept to each other, infinite number of images will be formed?

6.why a plain mirror is not used behind the light source in a searchlight?

Good questions.

Ask your teacher.

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