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anonymous

  • one year ago

Last question! YAY! How can I show that the normal line at any point on the circle x^2 + y^2 = r^2 passes through the origin?

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  1. dumbcow
    • one year ago
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    first show that slope of tangent line is dy/dx \[\frac{dy}{dx} = -\frac{x}{y}\] slope of normal is opp reciprocal \[m = \frac{y}{x}\] next use equation of normal line to show y-intercept is 0 \[y - y_0 = m(x - x_0)\] \[y = mx + (y_0 - mx_0)\] \[y = mx + (y_0 - \frac{y_0}{x_0} x_0) = mx\] Since y-intercept is 0, every normal line must go through origin

  2. anonymous
    • one year ago
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    Thank you! That makes sense now. :D

  3. dumbcow
    • one year ago
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    yw :)

  4. anonymous
    • one year ago
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    Oops! One last thing. How did you get -x/y for the tangent line?

  5. anonymous
    • one year ago
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    @dumbcow

  6. dumbcow
    • one year ago
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    taking the derivative of circle equation and solving for "dy/dx"

  7. anonymous
    • one year ago
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    Oh, sorry. Never mind. I think I figured it out.

  8. anonymous
    • one year ago
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    Thank you. :)

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