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Math Help - application problems

  1. #1
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    application problems

    I want to practice problems for applications of hyperbolas. I am having trouble finding a website so that I am able to practice these problems. Does anybody have a website they can recommend?

    The question I was given is like this?
    A curved mirror is placed in a store for a wide angle view of the room. the right hand branch of x squared over one minus y squared over three equals one models the curvature of the mirror. a small security camera is placed so that all of the 2-foot diameter of the mirror is visible. If the back of the room lies on x=18, what width of the back of the room is visible in the camera?
    Thank you
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  2. #2
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    Hello IDontunderstand
    Quote Originally Posted by IDontunderstand View Post
    I want to practice problems for applications of hyperbolas. I am having trouble finding a website so that I am able to practice these problems. Does anybody have a website they can recommend?

    The question I was given is like this?
    A curved mirror is placed in a store for a wide angle view of the room. the right hand branch of x squared over one minus y squared over three equals one models the curvature of the mirror. a small security camera is placed so that all of the 2-foot diameter of the mirror is visible. If the back of the room lies on x=18, what width of the back of the room is visible in the camera?
    Thank you
    I suggest you Google 'hyperbola examples' and see if you come up with anything that's useful to you.

    As far as this specific question is concerned, I think you need to clarify the details.

    First, if a wide angle view is required, then it must be a convex mirror. Therefore, if the mirror is represented by a segment of the right-hand branch of the hyperbola, then none of the wall at x = 18 will be visible, because the reflective surface of the mirror will be pointing in the negative x-direction.

    Next (on the assumption that the question in fact means the left-hand branch of the hyperbola) I think you will need to know the position of the camera in order to discover the visible width of the back wall. See the attached diagram. Unless you know a point through which the reflected ray must pass, you can't tell the direction of the incident ray.

    Try it with a small plane mirror. Stand a few feet away from a wall, with your back to the wall. Hold up the mirror at arm's length and study the width that's visible behind you. Now move the mirror closer to your eye: the closer your eye is to the mirror, the greater will be the visible width of the wall. Well, the same principle will apply even if the mirror is curved.

    So you need to get some clarification of what is required in this question.

    Grandad
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