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Mimi_x3

  • 3 years ago

At a point P on the parabola x^2=4ay, a normal PK is drawn. From the vertex O a perpendicular OM is drawn to meet the normal at M. Show that the equation of the locus of M as P varies on the parabola is x^4 -2ax^2y + x^2y^2 - ay^3 = 0

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  1. Ishaan94
    • 3 years ago
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    \[x^2 = 4ay\] |dw:1319018817978:dw| Differentiating with respect to x \[2x = 4a \frac{dy}{dx}\] \[\frac{x}{2a} =\frac{dy}{dx}\] Normal is perpendicular to tangent Hence, Slope of Normal is \(-\frac{2a}{x}\) @mimi can you draw the figure? I am confused

  2. Mimi_x3
    • 3 years ago
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    lols, im confused as well, i've attempted on drawing it but i think that it's wrong =/

  3. Ishaan94
    • 3 years ago
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    'From the vertex O a perpendicular OM is drawn to meet the normal at M' Maybe something is not right here, I am not sure but try reading the question again maybe some typo

  4. Ishaan94
    • 3 years ago
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    ah Okay I think I got it

  5. Ishaan94
    • 3 years ago
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    maybe the figure is like this|dw:1319019992221:dw|

  6. Kira_Yamato
    • 3 years ago
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    |dw:1319020480071:dw|

  7. Kira_Yamato
    • 3 years ago
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    |dw:1319020633975:dw|

  8. Kira_Yamato
    • 3 years ago
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    |dw:1319020681433:dw|

  9. Ishaan94
    • 3 years ago
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    equation of normal must be according to the y = mx + c form \[y' = \frac{-2a}{x}x' + c \]Now we have points (x,y) for the Normal on parabola \(x^2 = 4ay\) \[y = \frac{-2a}{x}x+c\] \[\frac{x^2}{4a} = -2a + c \] \[ \frac{x^2}{-8a^2} = c\]

  10. Ishaan94
    • 3 years ago
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    sorry, typo ignore the last expression

  11. Ishaan94
    • 3 years ago
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    \[\frac{x^2 + 2a*4a}{4a}=c\] \[\frac{x^2 + 8a^2}{4a}=c\]

  12. Ishaan94
    • 3 years ago
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    OM is from origin perpendicular to PK equation of OM \[y' = \frac{x}{2a}x'\]

  13. Ishaan94
    • 3 years ago
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    equation of PK \[y' = x' \frac{-2a}{x} + \frac{x^2 + 8a^2}{4a} \]

  14. Ishaan94
    • 3 years ago
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    The point M satisfies both the equation i.e equation of PK and equation of OM

  15. Ishaan94
    • 3 years ago
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    Let M be (x_1,y_1) \[y_1 = x_1* \frac{-2a}{x} + \frac{x^2 + 8a^2}{4a}\] \[y_! = x_1 *\frac{x}{2a}\] \[ {x_1*}\frac{-2a}{x} + \frac{x^2 + 8a^2}{4a} = \frac{x}{2a}*x_1\] \[ x_1 \left( \frac{4a^

  16. Ishaan94
    • 3 years ago
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    \[x_1 \left( \frac{4a^2 + x^2 }{2ax}\right) = \frac{x^2 + 8a^2}{4a}\]

  17. Ishaan94
    • 3 years ago
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    \[x_1 = \frac{x^3 + 8xa^2}{8a^2 + 2x^2}\]

  18. Ishaan94
    • 3 years ago
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    \[y_1 = \frac{x}{2a}\times \frac{x^3 + 8a^2x}{8a^2 +2x^2} \]

  19. Mimi_x3
    • 3 years ago
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    I'm lost =/ Isn't the equation of the normal y = -x + 2ap^3/p

  20. Ishaan94
    • 3 years ago
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    I don't think so \[ x^2 = 4ay \] \[\frac{dy}{dx} = \frac{x}{2a}\] Slope of tangent, Normal is perpendicular to tangent

  21. Ishaan94
    • 3 years ago
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    This is getting complicated, I will try this later I think I am getting the question wrong. I will try this on my notebook first

  22. Mimi_x3
    • 3 years ago
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    It's okay never mind then, I will ask my teacher tomorrow.

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